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2020 Paper 3 Q5
D: 1500.0 B: 1500.0
partial fractions integration polynomial identity differentiation improper integral limit evaluation

Show that for positive integer \(n\), \(x^n - y^n = (x-y)\displaystyle\sum_{r=1}^{n} x^{n-r} y^{r-1}\).

  1. Let \(\mathrm{F}\) be defined by \[ \mathrm{F}(x) = \frac{1}{x^n(x-k)} \quad \text{for } x \neq 0,\, k \] where \(n\) is a positive integer and \(k \neq 0\).
    1. Given that \[ \mathrm{F}(x) = \frac{A}{x-k} + \frac{\mathrm{f}(x)}{x^n}, \] where \(A\) is a constant and \(\mathrm{f}(x)\) is a polynomial, show that \[ \mathrm{f}(x) = \frac{1}{x-k}\left(1 - \left(\frac{x}{k}\right)^n\right). \] Deduce that \[ \mathrm{F}(x) = \frac{1}{k^n(x-k)} - \frac{1}{k}\sum_{r=1}^{n} \frac{1}{k^{n-r}x^r}. \]
    2. By differentiating \(x^n \mathrm{F}(x)\), prove that \[ \frac{1}{x^n(x-k)^2} = \frac{1}{k^n(x-k)^2} - \frac{n}{xk^n(x-k)} + \sum_{r=1}^{n} \frac{n-r}{k^{n+1-r}x^{r+1}}. \]
  2. Hence evaluate the limit of \[ \int_2^N \frac{1}{x^3(x-1)^2} \; \mathrm{d}x \] as \(N \to \infty\), justifying your answer.

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2016 Paper 1 Q1
D: 1500.0 B: 1516.0
polynomials polynomial identity factorisation algebraic manipulation binomial expansion geometric series

  1. For \(n=1\), \(2\), \(3\) and \(4\), the functions \(\p_n\) and \(\q_n\) are defined by \[ \p_n(x) = (x+1)^{2n} - (2n+1)x (x^2+x+1)^{n-1} \] and \[ \q_n(x) = \frac{x^{2n+1}+1}{x+1} \ \ \ \ \ \ \ \ \ \ \ \ (x\ne -1) \,. \ \ \ \ \ \ \ \ \ \ \] Show that \(\p_n(x)\equiv \q_n(x)\) (for \(x\ne-1\)) in the cases \(n=1\), \(n=2\) and \(n=3\). Show also that this does not hold in the case \(n=4\).
  2. Using results from part (i):
    • \(\bf (a)\) express \( \ \dfrac {300^3 +1}{301}\,\) as the product of two factors (neither of which is 1);
    • \(\bf (b)\) express \( \ \dfrac {7^{49}+1}{7^7+1}\,\) as the product of two factors (neither of which is 1), each written in terms of various powers of 7 which you should not attempt to calculate explicitly.

Solution:

  1. \(n=1\): \begin{align*} && p_1(x) &= (x+1)^2 - 3x(x^2+x+1)^0 \\ &&&= x^2+2x+1-3x \\ &&&= x^2-x+1\\ && q_1(x) &= \frac{x^3+1}{x+1} \\ &&&= x^2-x+1 = p_1(x) \\ \\ && p_2(x) &= (x+1)^4-5x(x^2+x+1)^1 \\ &&&= x^4+4x^3+6x^2+4x+1 - 5x^3-5x^2-5x \\ &&&= x^4-x^3+x^2-x+1 \\ &&q_2(x) &= \frac{x^5+1}{x+1} \\ &&&= x^4-x^3+x^2-x+1 = p_2(x) \\ \\ && p_3(x) &= (x+1)^6-7x(x^2+x+1)^2 \\ &&&= x^6+6x^5+15x^4+20x^3+15x^2+6x+1 - 7x(x^4+2x^3+3x^2+2x+1) \\ &&&= x^6-x^5+x^4-x^3+x^2-x+1 \\ && q_3(x) &= \frac{x^7+1}{x+1} \\ &&&= x^6-x^5+x^4-x^3+x^2-x+1 = p_3(x) \\ \\ && p_4(1) &= 2^8 - 9 \cdot 1 \cdot 3^3 \\ &&&= 256 - 243 = 13 \\ && q_4(1) &= \frac{2}{2} = 1 \neq 13 \end{align*}
    • \(\bf (a)\) \(\,\) \begin{align*} && \frac{300^3+1}{300+1} &= (300+1)^2 - 3 \cdot 300 \\ &&&= 301^2 - 30^2 \\ &&&= 271 \cdot 331 \end{align*}
    • \(\bf (b)\) \(\,\) \begin{align*} && \dfrac {7^{49}+1}{7^7+1} &= (7^7+1)^6 - 7 \cdot 7^7 \cdot (7^2+7+1)^2 \\ &&&= (7^7+1)^6 - 7^8 \cdot (7^2+7+1)^2 \\ &&&= ((7^7+1)^3 - 7^4(7^2+7+1)) \cdot ((7^7+1)^3 + 7^4(7^2+7+1)) \end{align*}

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2010 Paper 1 Q5
D: 1484.0 B: 1484.0
binomial theorem binomial coefficients summation identity substitution differentiation integration polynomial identity

By considering the expansion of \(\left(1+x\right)^{n}\) where \(n\) is a positive integer, or otherwise, show that:

  1. \[\binom{n}{0}+\binom{n}1+\binom{n}2 +\cdots +\binom{n}n=2^{n} \]
  2. \[\binom{n}{1}+2\binom{n}2+3\binom{n}3 +\cdots +n\binom{n}n=n2^{n-1} \]
  3. \[\binom{n}{0}+\frac12\binom{n}1+\frac13\binom{n}2 +\cdots +\frac1{n+1}\binom{n}n=\frac1{n+1}(2^{n+1}-1) \]
  4. \[\binom{n}{1}+2^2\binom{n}2+3^2\binom{n}3 +\cdots +n^2\binom{n}n=n(n+1)2^{n-2} \]

Solution:

  1. Notice that \begin{align*} && (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\ \text{Evaluate at }x = 1: && 2^n &= \sum_{i=0}^n \binom{n}{i} \end{align*}
  2. \(\,\) \begin{align*} && (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\ \frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{i=1}^n i\binom{n}{i} x^{i-1} \\ \text{Evaluate at }x = 1: && n2^{n-1} &= \sum_{i=1}^n i\binom{n}{i} \end{align*}
  3. \(\,\) \begin{align*} && (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\ \Rightarrow && \int_0^1(1+x)^n \d x &= \int_0^1 \sum_{i=0}^n \binom{n}{i} x^i \d x \\ \Rightarrow && \frac{1}{n+1}(2^{n+1}-1) &= \sum_{i=0}^n \binom{n}{i}\int_0^1 x^i \d x\\ &&& = \sum_{i=0}^n \frac{1}{i+1}\binom{n}{i} \\ \end{align*}
  4. \(\,\) \begin{align*} && (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\ \frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{i=1}^n i\binom{n}{i} x^{i-1} \\ \times x: && nx(1+x)^{n-1} &= \sum_{i=1}^n i\binom{n}{i} x^{i} \\ \frac{\d}{\d x}: && n(1+x)^{n-1}+n(n-1)x(1+x)^{n-2} &= \sum_{i=1}^n i^2\binom{n}{i} x^{i-1} \\ \text{Evaluate at }x = 1: && \sum_{i=1}^n i^2\binom{n}{i} &= n(1+1)^{n-1}+n(n-1)x(1+1)^{n-2} \\ &&&= 2^{n-2} \left (n(n-1) + 2n \right) \\ &&&= n(n+1)2^{n-2} \end{align*}

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2008 Paper 3 Q5
D: 1700.0 B: 1499.3
proof by induction recurrence relation sequences polynomial identity Chebyshev polynomials quadratic equation geometric progression

The functions \({\rm T}_n(x)\), for \(n=0\), 1, 2, \(\ldots\,\), satisfy the recurrence relation \[ {\rm T}_{n+1}(x) -2x {\rm T}_n(x) + {\rm T}_{n-1}(x) =0\, \ \ \ \ \ \ \ (n\ge1). \tag{\(*\)} \] Show by induction that \[ \left({\rm T}_n(x)\right)^2 - {\rm T}_{n-1}(x) {\rm T}_{n+1}(x) = \f(x)\,, \] where \(\f(x) = \left({\rm T}_1(x)\right)^2 - {\rm T}_0(x){\rm T}_2(x)\,\). In the case \(\f(x)\equiv 0\), determine (with proof) an expression for \({\rm T}_n(x)\) in terms of \({\rm T}_0(x)\) (assumed to be non-zero) and \({\rm r}(x)\), where \({\rm r}(x) = {\rm T}_1(x)/ {\rm T}_0(x)\). Find the two possible expressions for \({\rm r}(x)\) in terms of \(x\). %Conjecture (without proof) the general form of the solution of \((*)\).

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1995 Paper 1 Q3
D: 1500.0 B: 1500.0
sequences telescoping sum summation cube numbers method of differences alternating series polynomial identity

  1. If \(\mathrm{f}(r)\) is a function defined for \(r=0,1,2,3,\ldots,\) show that \[ \sum_{r=1}^{n}\left\{ \mathrm{f}(r)-\mathrm{f}(r-1)\right\} =\mathrm{f}(n)-\mathrm{f}(0). \]
  2. If \(\mathrm{f}(r)=r^{2}(r+1)^{2},\) evaluate \(\mathrm{f}(r)-\mathrm{f}(r-1)\) and hence determine \({\displaystyle \sum_{r=1}^{n}r^{3}.}\)
  3. Find the sum of the series \(1^{3}-2^{3}+3^{3}-4^{3}+\cdots+(2n+1)^{3}.\)

Solution:

  1. \(\,\) \begin{align*} && \sum_{r=1}^n \left (f(r) - f(r-1) \right) &= \cancel{f(1)} - f(0) + \cdots\\ &&&\quad\, \cancel{\f(2)}-\cancel{f(1)} + \cdots \\ &&&\quad\, \cancel{f(3)}-\cancel{f(2)} + \cdots \\ &&&\quad\, +\cdots + \cdots \\ &&&\quad\, \cancel{f(n-1)}-\cancel{f(n-2)} + \cdots \\ &&&\quad\, f(n)-\cancel{f(n-1)} \\ &&&=f(n) - f(0) \end{align*}
  2. If \(f(r) = r^2(r+1)^2\) then \begin{align*} && f(r) - f(r-1) &= r^2(r+1)^2 - (r-1)r^2 \\ &&&= r^2((r+1)^2-(r-1)^2) \\ &&&=4r^3 \end{align*} Therefore \begin{align*} && \sum_{r=1}^n 4r^3 &= n^2(n+1)^2-0 \\ \Rightarrow && \sum_{r=1}^n r^3 &= \frac{n^2(n+1)^2}{4} \\ \end{align*}
  3. So \begin{align*} && \sum_{r=1}^{2n+1} (-1)^{r-1}r^3 &= \sum_{r=1}^{2n+1} r^3 - 2 \sum_{r=1}^n (2r)^3 \\ &&&= \frac14(2n+1)^2(2n+2)^2 - 8 \frac{n^2(n+1)^2}{4} \\ &&&= (2n+1)^2(n+1)^2 - 2n^2(n+1)^2 \\ &&&= (n+1)^2(4n^2+4n+1 - 2n^2) \\ &&&= (n+1)^2(2n^2+4n+1) \end{align*}

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1995 Paper 2 Q1
D: 1600.0 B: 1484.0
geometric series differentiation alternating series polynomial identity summation substitution x=-1 second derivative

  1. By considering \((1+x+x^{2}+\cdots+x^{n})(1-x)\) show that, if \(x\neq1\), \[ 1+x+x^{2}+\cdots+x^{n}=\frac{1-x^{n+1}}{1-x}. \]
  2. By differentiating both sides and setting \(x=-1\) show that \[ 1-2+3-4+\cdots+(-1)^{n-1}n \] takes the value \(-n/2\) is \(n\) is even and the value \((n+1)/2\) if \(n\) is odd.
  3. Show that \[ 1^{2}-2^{2}+3^{2}-4^{2}+\cdots+(-1)^{n-1}n^{2}=(-1)^{n-1}(An^{2}+Bn) \] where the constants \(A\) and \(B\) are to be determined.

Solution:

  1. \begin{align*} && (1+x+x^{2}+\cdots+x^{n})(1-x) &= 1-x+x-x^2+\cdots -x^n+x^n-x^{n+1} \\ &&&= 1-x^{n+1} \\ \Rightarrow && 1+x+x^2+\cdots+x^n &= \frac{1-x^{n+1}}{1-x} \tag{dividing by \(1-x\)} \end{align*}
  2. \begin{align*} \frac{\d}{\d x}: && 0+1+2x+\cdots+nx^{n-1} &= \frac{(n+1)(1-x)x^n+(1-x^{n+1})}{(1-x)^2} \\ \Rightarrow && 1-2x+\cdots+(-1)^n n &= \frac{-(n+1)2(-1)^n+(1-(-1)^{n+1})}{4} \\ &&&= \begin{cases} \frac{-(n+1)\cdot2\cdot1+(1-(-1)}{4} & \text{if }n\text{ is even} \\ \frac{-(n+1)\cdot 2 \cdot(-1)+(1-1)}{4} & \text{if }n\text{ is odd}\end{cases} \\ &&&= \begin{cases} \frac{-n}{2} & \text{if }n\text{ is even}\\ \frac{n+1}{2} & \text{if }n\text{ is odd}\end{cases} \\ \end{align*}
  3. \begin{align*} x: && x+2x^2+\cdots+nx^{n} &= \frac{(n+1)(1-x)x^{n+1}+x(1-x^{n+1})}{(1-x)^2} \\ &&&= \frac{x+(n+1)x^{n+1}-nx^{n+2}}{(1-x)^2}\\ \frac{\d}{\d x}: && 1^2+2^2x + \cdots + n^2x^{n-1} &= \frac{(1-x)^2(1+(n+1)^2x^{n}-n(n+2)x^{n+1}) +2(1-x)(x+(n+1)x^{n+1}-nx^{n+2})}{(1-x)^4} \\ &&&= \frac{1 + x - (1 + n)^2 x^n + (2 n^2+2n-1) x^{n+1} - n^2 x^{n+2}}{(1-x)^3} \\ \Rightarrow && 1^2-2^2 + \cdots + (-1)^{n-1}n^2 &= \frac{(-1)^n \l - (1 + n)^2- (2 n^2+2n-1) - n^2 \r}{8} \\ &&&= \frac{(-1)^n(-4n^2-4n}{8} \\ &&&= \frac{(-1)^{n-1}(n^2+n)}{2} \end{align*}

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