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2025 Paper 2 Q4
D: 1500.0 B: 1500.0
proof by induction floor function integer part periodic function telescoping summation functional equation

Let \(\lfloor x \rfloor\) denote the largest integer that satisfies \(\lfloor x \rfloor \leq x\). For example, if \(x = -4.2\), then \(\lfloor x \rfloor = -5\).

  1. Show that, if \(n\) is an integer, then \(\lfloor x + n \rfloor = \lfloor x \rfloor + n\).
  2. Let \(n\) be a positive integer and define function \(f_n\) by \[f_n(x) = \lfloor x \rfloor + \left\lfloor x + \frac{1}{n} \right\rfloor + \left\lfloor x + \frac{2}{n} \right\rfloor + \ldots + \left\lfloor x + \frac{n-1}{n} \right\rfloor - \lfloor nx \rfloor\]
    1. Show that \(f_n\left(x + \frac{1}{n}\right) = f_n(x)\).
    2. Evaluate \(f_n(t)\) for \(0 \leq t < \frac{1}{n}\).
    3. Hence show that \(f_n(x) \equiv 0\).
    1. Show that \(\left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor = \lfloor x \rfloor\).
    2. Hence, or otherwise, simplify \[\left\lfloor \frac{x+1}{2} \right\rfloor + \left\lfloor \frac{x+2}{2^2} \right\rfloor + \ldots + \left\lfloor \frac{x+2^k}{2^{k+1}} \right\rfloor + \ldots\]

Solution:

  1. Claim: If \(n \in \mathbb{Z}\) then \(\lfloor x + n \rfloor = \lfloor x \rfloor + n\) Proof: Since \(\lfloor x \rfloor \leq x\) then \(\lfloor x \rfloor + n \leq x + n\) and \(\lfloor x \rfloor + n \in \mathbb{Z}\) we must have that \(\lfloor x \rfloor +n \leq \lfloor x + n \rfloor\). However, since \(\lfloor x \rfloor + 1 > x\) we must also have that \(\lfloor x \rfloor + 1 + n > x + n\), therefore \(\lfloor x \rfloor + n\) is the largest integer less than \(x + n\) as required.
    1. Claim: \(f_n\left(x + \frac{1}{n}\right) = f_n(x)\) Proof: \begin{align*} f_n\left(x + \frac{1}{n}\right) &=\left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{1}{n}+ \frac{1}{n} \right\rfloor + \left\lfloor x+ \frac{1}{n} + \frac{2}{n} \right\rfloor + \ldots + \left\lfloor x+ \frac{1}{n} + \frac{n-1}{n} \right\rfloor - \left \lfloor n\left ( x + \frac{1}{n} \right) \right \rfloor \\ &= \left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{2}{n}\right\rfloor + \left\lfloor x+ \frac{3}{n} \right\rfloor + \ldots + \left\lfloor x+ \frac{n}{n} \right\rfloor - \left \lfloor nx + 1 \right \rfloor \\ &= \left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{2}{n}\right\rfloor + \left\lfloor x+ \frac{3}{n} \right\rfloor + \ldots + \left\lfloor x+ 1 \right\rfloor - \left \lfloor nx + 1 \right \rfloor \\ &= \left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{2}{n}\right\rfloor + \left\lfloor x+ \frac{3}{n} \right\rfloor + \ldots + \lfloor x \rfloor + 1 - \left ( \lfloor nx \rfloor + 1 \right) \\ &= \lfloor x \rfloor + \left\lfloor x + \frac{1}{n} \right\rfloor + \left\lfloor x + \frac{2}{n} \right\rfloor + \ldots + \left\lfloor x + \frac{n-1}{n} \right\rfloor - \lfloor nx \rfloor \\ &= f_n(x) \end{align*}
    2. Suppose \(0 \leq t < \frac1n\), then note that \(\left \lfloor t + \frac{k}{n} \right \rfloor = 0\) for \(0 \leq k \leq n - 1\) and \(\lfloor n t \rfloor = 0\) since \(nt < 1\)
    3. Since \(f_n(x)\) is zero on \([0, \tfrac1n)\) and periodic with period \(\tfrac1n\) it must be constantly zero
    1. Claim: \(\left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor = \lfloor x \rfloor\) Proof: Suppose \(x = n + \epsilon\) where \(0 \leq \epsilon < 1\), ie \(n = \lfloor x \rfloor\), then consider two cases: Case 1: \(n = 2k\) \begin{align*} \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor &= \left\lfloor \frac{n + \epsilon}{2} \right\rfloor + \left\lfloor \frac{n + \epsilon+1}{2} \right\rfloor \\ &= \left\lfloor \frac{2k + \epsilon}{2} \right\rfloor + \left\lfloor \frac{2k + \epsilon+1}{2} \right\rfloor \\ &= k + \left\lfloor \frac{\epsilon}{2} \right\rfloor + k + \left\lfloor \frac{\epsilon+1}{2} \right\rfloor \\ &= 2k \\ &= n \end{align*} Case 2: \(n = 2k + 1\) \begin{align*} \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor &= \left\lfloor \frac{n + \epsilon}{2} \right\rfloor + \left\lfloor \frac{n + \epsilon+1}{2} \right\rfloor \\ &= \left\lfloor \frac{2k +1+ \epsilon}{2} \right\rfloor + \left\lfloor \frac{2k +1+ \epsilon+1}{2} \right\rfloor \\ &= k + \left\lfloor \frac{\epsilon+1}{2} \right\rfloor + k +1+ \left\lfloor \frac{\epsilon}{2} \right\rfloor \\ &= 2k +1\\ &= n \end{align*} as required.
    2. Since \(\left \lfloor \frac{x+1}{2} \right \rfloor = \lfloor x \rfloor - \lfloor \frac{x}{2} \rfloor\) and in general, \(\left \lfloor \frac{x+2^k}{2^{k+1}} \right \rfloor = \lfloor \frac{x}{2^k} \rfloor - \lfloor \frac{x}{2^{k+1}} \rfloor\) and so in general: \begin{align*} \sum_{k=0}^\infty \left \lfloor \frac{x+2^k}{2^{k+1}} \right \rfloor &= \sum_{k=0}^\infty \left ( \left \lfloor \frac{x}{2^k} \right \rfloor -\left \lfloor \frac{x}{2^{k+1}} \right \rfloor \right) \\ &= \lfloor x \rfloor \end{align*}

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2025 Paper 3 Q8
D: 1500.0 B: 1500.0
complex numbers proof by induction De Moivre trig identity summation sin nθ cos nθ algebraic identity

  1. Show that $$z^{m+1} - \frac{1}{z^{m+1}} = \left(z - \frac{1}{z}\right)\left(z^m + \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right)$$ Hence prove by induction that, for \(n \geq 1\), $$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$ Find similarly \(z^{2n} - \frac{1}{z^{2n}}\) as a product of \((z + \frac{1}{z})\) and a sum.
    1. By choosing \(z = e^{i\theta}\), show that $$\sin 2n\theta = 2\sin\theta \sum_{r=1}^n \cos(2r-1)\theta$$
    2. Use this result, with \(n = 2\), to show that \(\cos\frac{2\pi}{5} = \cos\frac{\pi}{5} - \frac{1}{2}\).
    3. Use this result, with \(n = 7\), to show that \(\cos\frac{2\pi}{15} + \cos\frac{4\pi}{15} + \cos\frac{8\pi}{15} + \cos\frac{16\pi}{15} = \frac{1}{2}\).
  3. Show that \(\sin\frac{\pi}{14} - \sin\frac{3\pi}{14} + \sin\frac{5\pi}{14} = \frac{1}{2}\).

Solution:

  1. \begin{align*} RHS &= \left(z - \frac{1}{z}\right)\left(z^m + \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right) \\ &= z^{m+1} + \frac{1}{z^{m-1}} - z^{m-1} - \frac{1}{z^{m+1}} + z^{m-1} - \frac{1}{z^{m-1}} \\ &= z^{m+1} - \frac{1}{z^{m+1}} \\ &= LHS \end{align*}. Claim: For \(n \geq 1\), $$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$ Proof: (By Induction) Base Case: (\(n=1\)). \begin{align*} LHS &= z^2 - \frac{1}{z^2} \\ &= (z-\frac1z)(z + \frac{1}{z}) \\ &= (z - \frac1z) \sum_{r=1}^1 \left ( z + \frac{1}{z} \right) \\ &= (z - \frac1z) \sum_{r=1}^1 \left ( z^{2r-1} + \frac{1}{z^{2r-1}} \right) \\ &= RHS \end{align*} as required. Inductive step: Suppose our result is true for some \(n=k\), then consider \(n = k+1\). \begin{align*} RHS &= \left(z - \frac{1}{z}\right)\sum_{r=1}^{k+1} \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right) \\ &= \left(z - \frac{1}{z}\right)\sum_{r=1}^{k} \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right) + \left(z - \frac{1}{z}\right)\left(z^{2k+1} + \frac{1}{z^{2k+1}}\right) \\ &= z^{2k} - \frac{1}{z^{2k}} + \left(z - \frac{1}{z}\right)\left(z^{2k+1} + \frac{1}{z^{2k+1}}\right) \\ &= z^{2k+2} - \frac{1}{z^{2k+2}} \\ &= LHS \end{align*}. Therefore if our result is true for \(n=k\) is true, it is true for \(n=k+1\). Since it is also true for \(n=1\) it is true for all \(n \geq 1\) but the principle of mathematical induction. Since \(\displaystyle z^{m+1} - \frac{1}{z^{m+1}} = \left(z + \frac{1}{z}\right)\left(z^m - \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right)\), we must have \(\displaystyle z^{2n}-\frac{1}{z^{2n}} = \left ( z + \frac{1}{z} \right) \sum_{r=1}^n \left (z^{2r-1}-\frac{1}{z^{2r-1}} \right)\)
    1. Since $$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$ we have \begin{align*} && e^{2n\theta i} - e^{-2n\theta i} &= \left(e^{\theta i} - e^{-\theta i}\right)\sum_{r=1}^n \left(e^{(2r-1)\theta i} + e^{-(2r-1)\theta i}\right) \\ \Rightarrow && 2i \sin 2n \theta &= 2i \sin \theta \sum_{r=1}^n 2 \cos (2r-1) \theta \\ \Rightarrow && \sin 2n \theta &= 2\sin \theta \sum_{r=1}^n \cos (2r-1) \theta \end{align*}
    2. When \(n = 2, \theta = \frac{\pi}{5}\) we have: \begin{align*} &&\sin \frac{4\pi}{5} &= 2 \sin \frac{\pi}{5} (\cos \frac{\pi}{5} + \cos \frac{3\pi}{5}) \\ &&\sin \frac{\pi}{5} &= 2 \sin \frac{\pi}{5} (\cos \frac{\pi}{5} - \cos \frac{2\pi}{5}) \\ &&\frac12 &= \cos \frac{\pi}{5} - \cos \frac{2 \pi}{5} \\ \Rightarrow && \cos \frac{2\pi}{5} &= \cos \frac{\pi}{5} - \frac12 \end{align*}
    3. When \(n = 7, \theta = \frac{\pi}{15}\) we have: \begin{align*} && \sin \frac{14 \pi}{15} &= 2 \sin \frac{\pi}{15} \sum_{r=1}^7 \cos (2r-1) \frac{\pi}{15} \\ \Rightarrow && \frac12 &= \cos \frac{\pi}{15} + \cos \frac{3 \pi}{15} + \cos \frac{5 \pi}{15}+ \cos \frac{7 \pi}{15}+ \cos \frac{9 \pi}{15}+ \cos \frac{11 \pi}{15}+ \cos \frac{13 \pi}{15} \\ &&&= -\cos \frac{16\pi}{15} + \cos \frac{3 \pi}{15} + \cos \frac{5 \pi}{15}- \cos \frac{8 \pi}{15}+ \cos \frac{9 \pi}{15}- \cos \frac{4 \pi}{15}- \cos \frac{2\pi}{15} \\ &&&= - \left ( \cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15}\right) + \cos \frac{\pi}{5} + \cos \frac{\pi}{3} + \cos \frac{3 \pi}{5} \\ &&&= - \left ( \cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15}\right) + \frac12 + \frac12 \\ \Rightarrow && \frac12 &= cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15} \end{align*}
  3. By using \(z = e^{i \theta}\) we have that: \begin{align*} && z^{2n}-\frac{1}{z^{2n}} &= \left ( z + \frac{1}{z} \right) \sum_{r=1}^n \left (z^{2r-1}-\frac{1}{z^{2r-1}} \right ) \\ \Rightarrow && e^{2n \theta i} - e^{-2n \theta i} &= (e^{\theta i} + e^{-\theta i}) \sum_{r=1}^n (e^{(2r-1)\theta i} - e^{(2r-1) \theta i}) \\ \Rightarrow && 2i \sin 2n \theta &= 2 \cos \theta \sum_{r=1}^n 2i \sin(2r-1) \theta \\ \Rightarrow && \sin 2n \theta &= 2 \cos \theta \sum_{r=1}^n \sin(2r-1) \theta \end{align*} When \(n = 3, \theta = \frac{\pi}{14}\) we must have: \begin{align*} &&\sin \frac{3 \pi}{7} &= 2 \cos \frac{\pi}{14}( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\ &&&= 2 \sin \left (\frac{\pi}{2} - \frac{\pi}{14} \right)( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\ &&&= 2 \sin \frac{3\pi}{7} ( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\ \Rightarrow && \frac12 &= \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14} \end{align*} as required.

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2024 Paper 2 Q6
D: 1500.0 B: 1500.0
generalised binomial theorem proof by induction recurrence relation binomial coefficients power series cauchy product

In this question, you need not consider issues of convergence.

  1. The sequence \(T_n\), for \(n = 0, 1, 2, \ldots\), is defined by \(T_0 = 1\) and, for \(n \geqslant 1\), by \[ T_n = \frac{2n-1}{2n}\,T_{n-1}. \] Prove by induction that \[ T_n = \frac{1}{2^{2n}}\binom{2n}{n}, \] for \(n = 0, 1, 2, \ldots\). [Note that \(\dbinom{0}{0} = 1\).]
  2. Show that in the binomial series for \((1-x)^{-\frac{1}{2}}\), \[ (1-x)^{-\frac{1}{2}} = \sum_{r=0}^{\infty} a_r x^r, \] successive coefficients are related by \[ a_r = \frac{2r-1}{2r}\,a_{r-1} \] for \(r = 1, 2, \ldots\)\,. Hence prove that \(a_r = T_r\) for all \(r = 0, 1, 2, \ldots\)\,.
  3. Let \(b_r\) be the coefficient of \(x^r\) in the binomial series for \((1-x)^{-\frac{3}{2}}\), so that \[ (1-x)^{-\frac{3}{2}} = \sum_{r=0}^{\infty} b_r x^r. \] By considering \(\dfrac{b_r}{a_r}\), find an expression involving a binomial coefficient for \(b_r\), for \(r = 0, 1, 2, \ldots\)\,.
  4. By considering the product of the binomial series for \((1-x)^{-\frac{1}{2}}\) and \((1-x)^{-1}\), prove that \[ \frac{(2n+1)}{2^{2n}}\binom{2n}{n} = \sum_{r=0}^{n} \frac{1}{2^{2r}}\binom{2r}{r}, \] for \(n = 1, 2, \ldots\)\,.

Solution:

  1. Claim: \(\displaystyle T_n = \frac{1}{2^{2n}}\binom{2n}{n}\) Proof: (By Induction) Base case: \(n=0\). Note that \(T_0 = 1\) and \(\frac{1}{2^0}\binom{0}{0} = 1\) so the base case is true. Assume true for some \(n=k\), ie \(T_k = \frac{1}{2^{2k}} \binom{2k}{k}\) so \begin{align*} && T_{k+1} &= \frac{2(k+1)-1}{2(k+1)} \frac{1}{2^{2k}} \binom{2k}{k} \\ &&&= \frac{2k+1}{k+1} \frac{1}{2^{2k+1}} \frac{(2k)!}{k!k!} \\ &&&= \frac{2(k+1)(2k+1)}{(k+1)(k+1)} \frac{1}{2^{2(k+1)}} \frac{(2k)!}{k!k!} \\ &&&= \frac{1}{2^{2(k+1)}} \frac{(2k+2)!}{(k+1)!(k+1)!} \\ &&&= \frac{1}{2^{2(k+1)}} \binom{2(k+1)}{k+1} \end{align*} and therefore it's true for all \(n\).
  2. Notice that \((1-x)^{-\frac12} = 1 + (-\tfrac12)(-x) + \frac{(-\frac12)(-\frac32)}{2!}(-x)^2+\cdots\) in particular \(a_r = \frac{(-\frac12 - r)}{r}(-1)a_{r-1} = \frac{2r-1}{2r}a_{r-1}\). Since \(a_0 = 1\) we have \(a_r = T_r\) for all \(r\).
  3. Notice that \begin{align*} && (1-x)^{-\frac32} &= \sum_{r=0}^\infty b_r x^r \\ &&&= \sum_{r=0}^\infty \frac{(-\frac32)\cdot(-\frac32-1)\cdots (-\frac32-(r-1))}{r!}(-x)^r \\ &&&= \sum_{r=0}^\infty \frac{(-\frac12-1)\cdot(-\frac12-2)\cdots (-\frac12-r)}{r!}(-x)^r \\ \end{align*} Therefore \(\frac{b_r}{a_r} = \frac{r+\frac12}{\frac12} = 2r+1\) so \(b_r = \frac{2r+1}{2^{2r}} \binom{2r}{r}\)
  4. Notice that \begin{align*} && (1-x)^{-\frac32} &= (1-x)^{-\frac12}(1-x)^{-1} \\ &&&= (1 + x+ x^2 + \cdots) \sum_{r=0}^{\infty} a_r x^r \\ &&&= \sum_{i=0}^{\infty} \sum_{k=0}^n a_r x^i \end{align*} So we must have \(b_r = \sum_{i=0}^ra_i\) which is the required result

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2023 Paper 2 Q5
D: 1500.0 B: 1500.0
recurrence relation sequences and series convergence inequality square root approximation newton-raphson method proof by induction

  1. The sequence \(x_n\) for \(n = 0, 1, 2, \ldots\) is defined by \(x_0 = 1\) and by \[x_{n+1} = \frac{x_n + 2}{x_n + 1}\] for \(n \geqslant 0\).
    1. Explain briefly why \(x_n \geqslant 1\) for all \(n\).
    2. Show that \(x_{n+1}^2 - 2\) and \(x_n^2 - 2\) have opposite sign, and that \[\left|x_{n+1}^2 - 2\right| \leqslant \tfrac{1}{4}\left|x_n^2 - 2\right|\,.\]
    3. Show that \[2 - 10^{-6} \leqslant x_{10}^2 \leqslant 2\,.\]
  2. The sequence \(y_n\) for \(n = 0, 1, 2, \ldots\) is defined by \(y_0 = 1\) and by \[y_{n+1} = \frac{y_n^2 + 2}{2y_n}\] for \(n \geqslant 0\).
    1. Show that, for \(n \geqslant 0\), \[y_{n+1} - \sqrt{2} = \frac{(y_n - \sqrt{2})^2}{2y_n}\] and deduce that \(y_n \geqslant 1\) for \(n \geqslant 0\).
    2. Show that \[y_n - \sqrt{2} \leqslant 2\left(\frac{\sqrt{2}-1}{2}\right)^{2^n}\] for \(n \geqslant 1\).
    3. Using the fact that \[\sqrt{2} - 1 < \tfrac{1}{2}\,,\] or otherwise, show that \[\sqrt{2} \leqslant y_{10} \leqslant \sqrt{2} + 10^{-600}\,.\]

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2023 Paper 2 Q6
D: 1500.0 B: 1500.0
proof by induction fibonacci sequence matrix exponentiation recurrence relation binomial theorem determinant 2x2 matrices

The sequence \(F_n\), for \(n = 0, 1, 2, \ldots\), is defined by \(F_0 = 0\), \(F_1 = 1\) and by \(F_{n+2} = F_{n+1} + F_n\) for \(n \geqslant 0\). Prove by induction that, for all positive integers \(n\), \[\begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix} = \mathbf{Q}^n,\] where the matrix \(\mathbf{Q}\) is given by \[\mathbf{Q} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}.\]

  1. By considering the matrix \(\mathbf{Q}^n\), show that \(F_{n+1}F_{n-1} - F_n^2 = (-1)^n\) for all positive integers \(n\).
  2. By considering the matrix \(\mathbf{Q}^{m+n}\), show that \(F_{m+n} = F_{m+1}F_n + F_m F_{n-1}\) for all positive integers \(m\) and \(n\).
  3. Show that \(\mathbf{Q}^2 = \mathbf{I} + \mathbf{Q}\). In the following parts, you may use without proof the Binomial Theorem for matrices: \[(\mathbf{I} + \mathbf{A})^n = \sum_{k=0}^{n} \binom{n}{k} \mathbf{A}^k.\]
    1. Show that, for all positive integers \(n\), \[F_{2n} = \sum_{k=0}^{n} \binom{n}{k} F_k\,.\]
    2. Show that, for all positive integers \(n\), \[F_{3n} = \sum_{k=0}^{n} \binom{n}{k} 2^k F_k\] and also that \[F_{3n} = \sum_{k=0}^{n} \binom{n}{k} F_{n+k}\,.\]
    3. Show that, for all positive integers \(n\), \[\sum_{k=0}^{n} (-1)^{n+k} \binom{n}{k} F_{n+k} = 0\,.\]

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2022 Paper 2 Q3
D: 1500.0 B: 1500.0
fibonacci numbers recurrence relation proof by induction sequences and series inequality decimal expansion telescoping sum

The Fibonacci numbers are defined by \(F_0 = 0\), \(F_1 = 1\) and, for \(n \geqslant 0\), \(F_{n+2} = F_{n+1} + F_n\).

  1. Prove that \(F_r \leqslant 2^{r-n} F_n\) for all \(n \geqslant 1\) and all \(r \geqslant n\).
  2. Let \(S_n = \displaystyle\sum_{r=1}^{n} \frac{F_r}{10^r}\). Show that \[\sum_{r=1}^{n} \frac{F_{r+1}}{10^{r-1}} - \sum_{r=1}^{n} \frac{F_r}{10^{r-1}} - \sum_{r=1}^{n} \frac{F_{r-1}}{10^{r-1}} = 89S_n - 10F_1 - F_0 + \frac{F_n}{10^n} + \frac{F_{n+1}}{10^{n-1}}\,.\]
  3. Show that \(\displaystyle\sum_{r=1}^{\infty} \frac{F_r}{10^r} = \frac{10}{89}\) and that \(\displaystyle\sum_{r=7}^{\infty} \frac{F_r}{10^r} < 2 \times 10^{-6}\). Hence find, with justification, the first six digits after the decimal point in the decimal expansion of \(\dfrac{1}{89}\).
  4. Find, with justification, a number of the form \(\dfrac{r}{s}\) with \(r\) and \(s\) both positive integers less than \(10000\) whose decimal expansion starts \[0.0001010203050813213455\ldots\,.\]

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2022 Paper 3 Q4
D: 1500.0 B: 1500.0
hyperbolic functions proof by induction infinite product maclaurin series limits logarithms trigonometric identities

You may assume that all infinite sums and products in this question converge.

  1. Prove by induction that for all positive integers \(n\), \[ \sinh x = 2^n \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right) \sinh\!\left(\frac{x}{2^n}\right) \] and deduce that, for \(x \neq 0\), \[ \frac{\sinh x}{x} \cdot \frac{\dfrac{x}{2^n}}{\sinh\!\left(\dfrac{x}{2^n}\right)} = \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right)\,. \]
  2. You are given that the Maclaurin series for \(\sinh x\) is \[ \sinh x = \sum_{r=0}^{\infty} \frac{x^{2r+1}}{(2r+1)!}\,. \] Use this result to show that, as \(y\) tends to \(0\), \(\dfrac{y}{\sinh y}\) tends to \(1\). Deduce that, for \(x \neq 0\), \[ \frac{\sinh x}{x} = \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right) \cdots\,. \]
  3. Let \(x = \ln 2\). Evaluate \(\cosh\!\left(\dfrac{x}{2}\right)\) and show that \[ \cosh\!\left(\frac{x}{4}\right) = \frac{1 + 2^{\frac{1}{2}}}{2 \times 2^{\frac{1}{4}}}\,. \] Use part (ii) to show that \[ \frac{1}{\ln 2} = \frac{1 + 2^{\frac{1}{2}}}{2} \times \frac{1 + 2^{\frac{1}{4}}}{2} \times \frac{1 + 2^{\frac{1}{8}}}{2} \cdots\,. \]
  4. Show that \[ \frac{2}{\pi} = \frac{\sqrt{2}}{2} \times \frac{\sqrt{2+\sqrt{2}}}{2} \times \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2} \cdots\,. \]

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2021 Paper 2 Q8
D: 1500.0 B: 1500.0
integration by parts recurrence relation differentiation proof by induction irrationality sequences and series exponential function

  1. Show that, for \(n = 2, 3, 4, \ldots\), \[ \frac{d^2}{dt^2}\bigl[t^n(1-t)^n\bigr] = n\,t^{n-2}(1-t)^{n-2}\bigl[(n-1) - 2(2n-1)t(1-t)\bigr]. \]
  2. The sequence \(T_0, T_1, \ldots\) is defined by \[ T_n = \int_0^1 \frac{t^n(1-t)^n}{n!}\,e^t\,dt. \] Show that, for \(n \geqslant 2\), \[ T_n = T_{n-2} - 2(2n-1)T_{n-1}. \]
  3. Evaluate \(T_0\) and \(T_1\) and deduce that, for \(n \geqslant 0\), \(T_n\) can be written in the form \[ T_n = a_n + b_n e, \] where \(a_n\) and \(b_n\) are integers (which you should not attempt to evaluate).
  4. Show that \(0 < T_n < \dfrac{e}{n!}\) for \(n \geqslant 0\). Given that \(b_n\) is non-zero for all~\(n\), deduce that \(\dfrac{-a_n}{b_n}\) tends to \(e\) as \(n\) tends to infinity.

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2021 Paper 2 Q11
D: 1500.0 B: 1500.0
conditional probability recurrence relation proof by induction combinatorics random seating problem discrete probability

A train has \(n\) seats, where \(n \geqslant 2\). For a particular journey, all \(n\) seats have been sold, and each of the \(n\) passengers has been allocated a seat. The passengers arrive one at a time and are labelled \(T_1, \ldots, T_n\) according to the order in which they arrive: \(T_1\) arrives first and \(T_n\) arrives last. The seat allocated to \(T_r\) (\(r = 1, \ldots, n\)) is labelled \(S_r\). Passenger \(T_1\) ignores their allocation and decides to choose a seat at random (each of the \(n\) seats being equally likely). However, for each \(r \geqslant 2\), passenger \(T_r\) sits in \(S_r\) if it is available or, if \(S_r\) is not available, chooses from the available seats at random.

  1. Let \(P_n\) be the probability that, in a train with \(n\) seats, \(T_n\) sits in \(S_n\). Write down the value of \(P_2\) and find the value of \(P_3\).
  2. Explain why, for \(k = 2, 3, \ldots, n-1\), \[ \mathrm{P}\bigl(T_n \text{ sits in } S_n \mid T_1 \text{ sits in } S_k\bigr) = P_{n-k+1}, \] and deduce that, for \(n \geqslant 3\), \[ P_n = \frac{1}{n}\Biggl(1 + \sum_{r=2}^{n-1} P_r\Biggr). \]
  3. Give the value of \(P_n\) in its simplest form and prove your result by induction.
  4. Let \(Q_n\) be the probability that, in a train with \(n\) seats, \(T_{n-1}\) sits in \(S_{n-1}\). Determine \(Q_n\) for \(n \geqslant 2\).

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2021 Paper 3 Q8
D: 1500.0 B: 1500.0
recurrence relation sequences and series proof by induction convergence inequality telescoping product

A sequence \(x_1, x_2, \ldots\) of real numbers is defined by \(x_{n+1} = x_n^2 - 2\) for \(n \geqslant 1\) and \(x_1 = a\).

  1. Show that if \(a > 2\) then \(x_n \geqslant 2 + 4^{n-1}(a-2)\).
  2. Show also that \(x_n \to \infty\) as \(n \to \infty\) if and only if \(|a| > 2\).
  3. When \(a > 2\), a second sequence \(y_1, y_2, \ldots\) is defined by \[ y_n = \frac{Ax_1 x_2 \cdots x_n}{x_{n+1}}, \] where \(A\) is a positive constant and \(n \geqslant 1\). Prove that, for a certain value of \(a\), with \(a > 2\), which you should find in terms of \(A\), \[ y_n = \frac{\sqrt{x_{n+1}^2 - 4}}{x_{n+1}} \] for all \(n \geqslant 1\). Determine whether, for this value of \(a\), the second sequence converges.

Solution:

  1. Claim \(x_n \geqslant 2 + 4^{n-1}(a-2)\) Proof: (By induction) Base case: Note that when \(n = 1\), \(x_1 = a = 2 + 1 \cdot(a - 2)\). Inductive step, suppose true for some \(n\), then \begin{align*} && x_{n+1} &= x_n^2 - 2 \\ &&&\geq (2+4^{n-1}(a-2))^2 - 2 \\ &&&= 4 + 4^{2n-2}(a-2)^2 + 4^n(a-2) - 2 \\ &&&= 2 + 4^{n}(a-2) + 4^{2n-2}(a-2)^2 \\ &&&\geq 2 + 4^{n+1-1}(a-2) \end{align*} as required,
  2. (\(\Leftarrow\)) Suppose \(a > 2\) then \(x_n \geq 2+4^{n-1}(a-2) \to \infty\) as required. Suppose \(a < -2\) then \(x_2 > 4 -2 = 2\) so the sequence starting from \(x_2\) clearly diverges for the same reason. (\(\Rightarrow\)) suppose \(|x_n| \leq 2\) then \(x_{n+1} = x_n^2 - 2 \leq 2\) so the sequence is bounded and cannot tend to \(\infty\).
  3. Suppose \(y_n = \frac{Ax_1x_2 \cdots x_n}{x_{n+1}}\) and notice that \(x_{n+1}^2 - 4 = (x_n^2 -2)^2 - 4 = x_n^4 - 4x_n^2 = x_n^2(x_n^2-4)\). In particular, \(\frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}} = \frac{x_n\sqrt{x_n^2-4}}{x_{n+1}} = \frac{x_n x_{n-1} \cdots x_1 \sqrt{x_1^2-4}}{x_{n+1}}\) Therefore if \(A = \sqrt{a^2-4}\) \(y_{n+1} = \frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}}\). Notice that \begin{align*} && y_n &= \frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}} \\ &&&= \sqrt{1 - \frac{4}{x_{n+1}^2}} \to 1 \end{align*}

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2020 Paper 2 Q3
D: 1500.0 B: 1500.0
sequences and series unimodal sequence recurrence relation proof by induction inequality logarithmic concavity

A sequence \(u_1, u_2, \ldots, u_n\) of positive real numbers is said to be unimodal if there is a value \(k\) such that \[u_1 \leqslant u_2 \leqslant \ldots \leqslant u_k\] and \[u_k \geqslant u_{k+1} \geqslant \ldots \geqslant u_n.\] So the sequences \(1, 2, 3, 2, 1\);\ \(1, 2, 3, 4, 5\);\ \(1, 1, 3, 3, 2\) and \(2, 2, 2, 2, 2\) are all unimodal, but \(1, 2, 1, 3, 1\) is not. A sequence \(u_1, u_2, \ldots, u_n\) of positive real numbers is said to have property \(L\) if \(u_{r-1}u_{r+1} \leqslant u_r^2\) for all \(r\) with \(2 \leqslant r \leqslant n-1\).

  1. Show that, in any sequence of positive real numbers with property \(L\), \[u_{r-1} \geqslant u_r \implies u_r \geqslant u_{r+1}.\] Prove that any sequence of positive real numbers with property \(L\) is unimodal.
  2. A sequence \(u_1, u_2, \ldots, u_n\) of real numbers satisfies \(u_r = 2\alpha u_{r-1} - \alpha^2 u_{r-2}\) for \(3 \leqslant r \leqslant n\), where \(\alpha\) is a positive real constant. Prove that, for \(2 \leqslant r \leqslant n\), \[u_r - \alpha u_{r-1} = \alpha^{r-2}(u_2 - \alpha u_1)\] and, for \(2 \leqslant r \leqslant n-1\), \[u_r^2 - u_{r-1}u_{r+1} = (u_r - \alpha u_{r-1})^2.\] Hence show that the sequence consists of positive terms and is unimodal, provided \(u_2 > \alpha u_1 > 0\). In the case \(u_1 = 1\) and \(u_2 = 2\), prove by induction that \(u_r = (2-r)\alpha^{r-1} + 2(r-1)\alpha^{r-2}\). Let \(\alpha = 1 - \dfrac{1}{N}\), where \(N\) is an integer with \(2 \leqslant N \leqslant n\). In the case \(u_1 = 1\) and \(u_2 = 2\), prove that \(u_r\) is largest when \(r = N\).

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2020 Paper 3 Q1
D: 1500.0 B: 1500.0
reduction formulae integration by parts proof by induction trigonometric identities definite integral recurrence relation

For non-negative integers \(a\) and \(b\), let \[ \mathrm{I}(a,b) = \int_0^{\frac{\pi}{2}} \cos^a x \cos bx \; \mathrm{d}x. \]

  1. Show that for positive integers \(a\) and \(b\), \[ \mathrm{I}(a,b) = \frac{a}{a+b} \, \mathrm{I}(a-1, b-1). \]
  2. Prove by induction on \(n\) that for non-negative integers \(n\) and \(m\), \[ \int_0^{\frac{\pi}{2}} \cos^n x \cos(n+2m+1)x \; \mathrm{d}x = (-1)^m \frac{2^n \, n! \, (2m)! \, (n+m)!}{m! \, (2n+2m+1)!}. \]

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2019 Paper 2 Q3
D: 1500.0 B: 1500.0
polynomials triangle inequality proof by induction roots of polynomials inequality geometric series bound on roots rational root theorem

For any two real numbers \(x_1\) and \(x_2\), show that $$|x_1 + x_2| \leq |x_1| + |x_2|.$$ Show further that, for any real numbers \(x_1, x_2, \ldots, x_n\), $$|x_1 + x_2 + \cdots + x_n| \leq |x_1| + |x_2| + \cdots + |x_n|.$$

  1. The polynomial f is defined by $$f(x) = 1 + a_1 x + a_2 x^2 + \cdots + a_{n-1} x^{n-1} + x^n$$ where the coefficients are real and satisfy \(|a_i| \leq A\) for \(i = 1, 2, \ldots, n-1\), where \(A \geq 1\).
    1. If \(|x| < 1\), show that $$|f(x) - 1| \leq \frac{A|x|}{1 - |x|}.$$
    2. Let \(\omega\) be a real root of f, so that \(f(\omega) = 0\). In the case \(|\omega| < 1\), show that $$\frac{1}{1 + A} \leq |\omega| \leq 1 + A. \quad (*)$$
    3. Show further that the inequalities \((*)\) also hold if \(|\omega| \geq 1\).
  2. Find the integer root or roots of the quintic equation $$135x^5 - 135x^4 - 100x^3 - 91x^2 - 126x + 135 = 0.$$

Solution: Claim: \(|x_1 + x_2| \leq |x_1| + |x_2|\) Proof: Case 1: \(x_1, x_2 \geq 0\). The inequality is equivalent to \(|x_1 + x_2| = x_1 + x_2 = |x_1|+|x_2|\) so it's an equality. Case 2: \(x_1, x_2 \leq 0\). The inequality is equivalent to \(|x_1+x_2| = -x_1-x_2 = |x_1|+|x_2\), so it's also an equality in this case. Case 3: (wlog) \(|x_1| \geq |x_2| > 0\) and \(x_1x_2 < 0\) then \(|x_1+x_2| = x_1-x_2 \leq x_1 \leq |x_1|+|x_2|\) We can prove this by induction, we've already proven the base case and: \(|x_1+x_2 + \cdots + x_n| \leq |x_1 + x_2 + \cdots x_{n-1}| + |x_n| \leq |x_1| + |x_2| + \cdots + |x_n|\)

  1. \(\,\) \begin{align*} && |f(x) - 1| &= |a_1 x + a_2x^2 + \cdots + a_{n-1}x^{n-1} + x^n| \\ &&&\leq |a_1x| + |a_2x^2| + \cdots + |a_{n-1}x^{n-1}| + |x^n| \\ &&&\leq |a_1||x| + |a_2||x|^2 + \cdots + |a_{n-1}||x|^{n-1} + |x|^n \\ &&&\leq A|x| + A|x|^2 + \cdots + A|x|^{n-1} + |x|^n \\ &&&=A|x| \frac{1-|x|^{n-1}}{1-|x|} + |x|^n \\ &&&= \frac{A|x|-A|x|^{n}+|x|^{n+1}-|x|^n}{1-|x|} \\ &&&= \frac{A|x|-|x|^n(\underbrace{A-|x|+1}_{\geq0})}{1-|x|} \\ &&&\leq \frac{A|x|}{1-|x|} \end{align*}
  2. If \(f(\omega) = 0\) then \begin{align*} && 1 & \leq \frac{A|\omega|}{1-|\omega|} \\ \Leftrightarrow && 1-|\omega| &\leq A |\omega| \\ \Leftrightarrow && 1 &\leq (1+A) |\omega| \\ \Leftrightarrow && \frac{1}{1+A} &\leq |\omega| \\ \end{align*} We also know \(\omega \leq 1 < 1 + A\)
  3. If \(\omega\) is a root of \(f(x)\) then \(1/\omega\) is a root of \(1 + a_{n-1}x + a_{n-2}x^2 + \cdots + a_1x^{n-1}+x^n\) and so \(1/\omega\) satisfies that inequality, ie \begin{align*} && \frac{1}{1+A} && \leq &&|1/\omega| && \leq &&1 + A \\ \Leftrightarrow &&1+A && \geq&& |\omega| && \geq&& \frac{1}{1 + A} \end{align*}
  4. First notice that it's equivalent to: \(0 = x^5 - 1x^4 - \frac{100}{135}x^3-\frac{91}{135}x^2-\frac{126}{135} + 1\) therefore all integer roots must be between \(-2,-1\) and \(1\) and \(2\). \(1\) doesn't work. \(-1\) works. Clearly \(2\) cannot work by parity argument, therefore the only integer root is \(-1\).

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2018 Paper 3 Q2
D: 1700.0 B: 1516.0
differentiation sequences recurrence relation proof by induction Gaussian function inequality Hermite polynomials product rule

The sequence of functions \(y_0\), \(y_1\), \(y_2\), \(\ldots\,\) is defined by \(y_0=1\) and, for \(n\ge1\,\), \[ y_n = (-1)^n \frac {1}{z} \, \frac{\d^{n} z}{\d x^n} \,, \] where \(z= \e^{-x^2}\!\).

  1. Show that \(\dfrac{\d y_n}{\hspace{-4.7pt}\d x} = 2x y_n -y_{n+1}\,\) for \(n\ge1\,\).
  2. Prove by induction that, for \(n\ge1\,\), \[ y_{n+1} = 2x y_n -2ny_{n-1} \,. \] Deduce that, for \(n\ge1\,\), \[ y_{n+1}^2 - {y}_n {y}_{n+2} = 2n (y_n^2 - y_{n-1}y_{n+1}) + 2 y_n^2 \,. \]
  3. Hence show that $y_{n}^2 - y^2_{n-1} y^2_{n+1} > 0\( for \)n \ge 1$.

Solution:

  1. \begin{align*} \frac{\d y_n}{\d x} &= \frac{\d}{\d x} \l (-1)^n e^{x^2} \frac{\d^n}{\d x^{n}} \l e^{-x^2}\r \r \\ &= (-1)^n 2xe^{x^2} \frac{\d^n}{\d x^{n}} \l e^{-x^2}\r + (-1)^n e^{x^2} \frac{\d^{n+1}}{\d x^{n+1}} \l e^{-x^2}\r \\ &= 2xy_n - (-1)^{n+1} e^{x^2} \frac{\d^{n+1}}{\d x^{n+1}} \l e^{-x^2}\r \\ &= 2xy_n - y_{n+1} \end{align*}
  2. \(y_0 = 1\), \(y_1 = (-1) e^{x^2} \cdot (-2x) \cdot e^{-x^2} = 2x\), \(y_2 = e^{x^2} \frac{\d^2}{\d x^2} \l e^{-x^2}\r = e^{x^2} \frac{\d }{\d x}\l -2xe^{-x^2} \r = e^{x^2} \l -2e^{-x^2}+4x^2e^{-x^2}\r = 4x^2-2\). Therefore \(2xy_1 - 2y_0 = 2x \cdot 2x - 2\cdot1 = 4x^2-2 = y_2\) so our statement is true for \(n=1\). Assume the statement is true for \(n=k\), then \begin{align*} && y_{k+1} &= 2xy_k - 2ky_{k-1} \\ \frac{\d }{\d x}: && \frac{\d y_{k+1}}{\d x} &= 2\frac{\d}{\d x}\l xy_k \r - 2k\frac{\d y_{k-1}}{\d x} \\ \Rightarrow && 2xy_{k+1}-y_{k+2} &= 2y_k+2x \l 2xy_k-y_{k+1}\r - 2k \l 2xy_{k-1}-y_k \r \\ \Rightarrow && y_{k+2} &=2y_k+ 4x \cdot y_{k+1}-(4x^2+2k)y_k+2x \cdot 2k y_{k-1} \\ &&&= 4x \cdot y_{k+1}-(4x^2+2(k+1))y_k+2x \l2xy_k - y_{k+1} \r \\ &&&= 2x \cdot y_{k+1} -2(k+1)y_k \end{align*} Therefore since our statement is true for \(n=1\) and if it is true for \(n=k\) it is true for \(n=k=1\), therefore by the principle of mathematical induction it is true for all \(n \geq 1\). Since \(2x = \frac{y_{n+1}+2ny_{n-1}}{y_n}\) for all \(n\), we must have \begin{align*} && \frac{y_{n+1}+2ny_{n-1}}{y_n} &= \frac{y_{n+2}+2(n+1)y_{n}}{y_{n+1}} \\ \Leftrightarrow && y_{n+1}^2+2ny_{n-1}y_{n+1} &= y_ny_{n+2}+2ny_n^2+2y_n^2 \\ \Leftrightarrow && y_{n+1}^2-y_ny_{n+2} &= 2n(y_n^2-y_{n-1}y_{n+1})+2y_n^2 \end{align*}
  3. Consider the functions \(f_n(x) = y_{n}^2-y_{n-1}y_{n+1}\) then clearly \(f_{n+1} = 2nf_{n} + 2y_n^2 \geq f_{n}\) so to prove \(f_n(x) > 0\) for \(n \geq 1\) it suffices to prove it for \(n = 1\). But \(f_1 = y_1^2 - y_0y_{2} = (2x)^2-(4x^2-2) = 2 > 0\) so we are done.

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2018 Paper 3 Q5
D: 1700.0 B: 1484.0
inequality AM-GM inequality sequences differentiation from first principles proof by induction polynomial analysis stationary points optimisation

The real numbers \(a_1\), \(a_2\), \(a_3\), \(\ldots\) are all positive. For each positive integer \(n\), \(A_n\) and \(G_n\) are defined by \[ A_n = \frac{a_1+a_2 + \cdots + a_n}n \ \ \ \ \ \text{and } \ \ \ \ \ G_n = \big( a_1a_2\cdots a_n\big) ^{1/n} \,. \]

  1. Show that, for any given positive integer \(k\), \[ (k+1) ( A_{k+1} - G_{k+1}) \ge k (A_k-G_k) \] if and only if \[\lambda^{k+1}_k -(k+1)\lambda_{{k}} +k \ge 0\,, \] where \( \lambda_{{k}} = \left(\dfrac{a_{k+1}}{G_{k}}\right)^{\frac1 {k+1}}\,\).
  2. Let \[ \f(x)=x^{k+1} -(k+1)x +k \,, \] where \(x > 0\) and \(k\) is a positive integer. Show that \(\f(x)\ge0\) and that \(\f(x)=0\) if and only if \(x = 1\,\).
  3. Deduce that:
    1. \(A_n \ge G_n\) for all \(n\); \\
    2. if \(A_n=G_n\) for some \(n\), then \(a_1=a_2 = \cdots = a_n\,\).

Solution:

  1. \begin{align*} && (k+1) (A_{k+1} - G_{k+1}) & \geq k(A_k - G_k) \\ \Leftrightarrow && \sum_{i=1}^{k+1} a_i - (k+1)G_{k+1} &\geq \sum_{i=1}^k a_i - kG_k \\ \Leftrightarrow && a_{k+1} -(k+1)G_k^{k/(k+1)}a_{k+1}^{1/(k+1)} & \geq - k G_k \\ \Leftrightarrow && a_{k+1} -(k+1)G_k^{k/(k+1)}a_{k+1}^{1/(k+1)} + k G_k& \geq 0\\ \Leftrightarrow && \frac{a_{k+1}}{G_k} -(k+1)G_k^{k/(k+1)-1}a_{k+1}^{1/(k+1)} + k & \geq 0\\ \Leftrightarrow && \lambda_k^{k+1} -(k+1)\lambda_k+ k & \geq 0\\ \end{align*} as required.
  2. \begin{align*} && f'(x) &= (k+1)x^k - (k+1) \\ &&&= (k+1)(x^k-1) \end{align*} Therefore \(f(x)\) is strictly decreasing on \((0,1)\) and strictly increasing on \((1,\infty)\) and so the minimum will be \(f(1) = 1 - (k+1) + k = 0\), so \(f(x) \geq 0\) with equality only at \(x = 1\).
    1. We can proceed by induction to show since the inequality holds for \(n=1\) and since if it holds for \(n=k\) it will hold for \(n=k+1\) as \(A_{k+1}-G_{k+1}\) must have the same sign as \(A_k - G_k\).
    2. The only way for equality to hold is if \(\lambda_k = 1\) for \(k = 1, \cdots n\), ie \(a_{k+1} = G_k\), but this means \(a_2 = a_1, a_3 = a_1\) etc. Therefore all values are equal.

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