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2024 Paper 2 Q2
D: 1500.0 B: 1500.0
generalised binomial theorem integration partial fractions series expansion term-by-term integration logarithm trigonometric identities

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

  1. Find the binomial series expansion of \((8 + x^3)^{-1}\), valid for \(|x| < 2\). Hence show that, for each integer \(m \geqslant 0\), \[ \int_0^1 \frac{x^m}{8 + x^3}\,\mathrm{d}x = \sum_{k=0}^{\infty} \left( \frac{(-1)^k}{2^{3(k+1)}} \cdot \frac{1}{3k + m + 1} \right). \]
  2. Show that \[ \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{1}{3k+3} - \frac{2}{3k+2} + \frac{4}{3k+1} \right) = \int_0^1 \frac{1}{x+2}\,\mathrm{d}x\,, \] and evaluate the integral.
  3. Show that \[ \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{72(2k+1)}{(3k+1)(3k+2)} \right) = \pi\sqrt{a} - \ln b\,, \] where \(a\) and \(b\) are integers which you should determine.

Solution:

  1. Note that \(\,\) \begin{align*} && (8+x^3)^{-1} &= \tfrac18(1 + \tfrac18x^3)^{-1} \\ &&&= \tfrac18( 1 - \left (\tfrac{x}{2} \right)^3 + \left (\tfrac{x}{2} \right)^6 -\left (\tfrac{x}{2} \right)^9 + \cdots ) \end{align*} So \begin{align*} && \int_0^1 \frac{x^m}{8+x^3} \d x &= \int_0^1 x^m \sum_{k=0}^{\infty} \frac{1}{2^3} \left ( - \frac{x}{2} \right)^{3k} \d x \\ &&&= \sum_{k=0}^{\infty} (-1)^k \frac{1}{2^{3(k+1)}} \int_0^1 x^{m+3k} \d x \\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \frac{1}{3k+m+1} \\ \end{align*}
  2. Notice that \begin{align*} && S &= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{1}{3k+3} - \frac{2}{3k+2} + \frac{4}{3k+1} \right) \\ &&&= \int_0^1 \frac{x^2}{8+x^3} \d x - 2\int_0^1 \frac{x^1}{8+x^3}+4\int_0^1 \frac{x^0}{8+x^3} \d x \\ &&&= \int_0^1 \frac{x^2-2x+4}{x^3+8} \d x \\ &&&= \int_0^1 \frac{1}{x+2} \d x = \left[ \ln(x+2)\right]_0^1 \\ &&&= \ln 3 - \ln 2 = \ln \tfrac32 \end{align*}
  3. Firstly, note that \begin{align*} && \frac{2k+1}{(3k+1)(3k+2)} &= \frac13 \left ( \frac{1}{3k+1} +\frac{1}{3k+2} \right) \end{align*} so \begin{align*} && S_2 &= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{72(2k+1)}{(3k+1)(3k+2)} \right) \\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} 24 \left ( \frac{1}{3k+1} +\frac{1}{3k+2} \right) \\ &&&= 24 \left [ \int_0^1 \frac{x^0}{8+x^3} \d x + \int_0^1 \frac{x^1}{8+x^3} \d x \right] \\ &&&= 24 \left [ \int_0^1 \frac{1+x}{(x+2)(x^2-2x+4)} \d x \right] \\ &&&= 24 \left [ \int_0^1 \frac{x+8}{12(x^2-2x+4)}-\frac{1}{12(x+2)} \d x \right] \\ &&&= 2\left [ \int_0^1 \frac{x+8}{x^2-2x+4}-\frac{1}{x+2} \d x \right] \\ &&&= 2\left [ \int_0^1 \frac{x-1}{x^2-2x+4}+ \frac{9}{(x-1)^2+3}-\frac{1}{x+2} \d x \right] \\ &&&=2 \left [ \int_0^1 \frac12\ln(x^2-2x+4)+3\sqrt{3}\arctan \frac{x-1}{\sqrt{3}} -\ln(x+2) \right]_0^1 \\ &&&=2 \left ( \frac12 \ln3+3\sqrt3 \arctan 0 - \ln 3 \right) - 2\left (\frac12 \ln 4-3\sqrt3 \arctan \frac{1}{\sqrt3} - \ln 2 \right) \\ &&&=\sqrt3 \pi - \ln3 \end{align*}

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2014 Paper 3 Q1
D: 1700.0 B: 1542.2
roots of polynomials symmetric functions Newton's sums logarithmic series power sums series expansion proof by counterexample

Let \(a\), \(b\) and \(c\) be real numbers such that \(a+b+c=0\) and let \[(1+ax)(1+bx)(1+cx) = 1+qx^2 +rx^3\,\] for all real \(x\). Show that \(q = bc+ca+ab\) and \(r= abc\).

  1. Show that the coefficient of \(x^n\) in the series expansion (in ascending powers of \(x\)) of \(\ln (1+qx^2+rx^3)\) is \((-1)^{n+1} S_n\) where \[S_n = \frac{a^n+b^n+c^n}{n} \,, \ \ \ \ \ \ \ \ (n\ge1).\]
  2. Find, in terms of \(q\) and \(r\), the coefficients of \(x^2\), \(x^3\) and \(x^5\) in the series expansion (in ascending powers of \(x\)) of \(\ln (1+qx^2+rx^3)\) and hence show that \(S_2S_3 =S_5\).
  3. Show that \(S_2S_5 =S_7\).
  4. Give a proof of, or find a counterexample to, the claim that \(S_2S_7=S_9\).

Solution: \begin{align*} (1+ax)(1+bx)(1+cx) &= (1+(a+b)x+abx^2)(1+cx) \\ &= 1+(a+b+c)x+(ab+bc+ca)x^2+abcx^3 \end{align*} Therefore by comparing coefficients, \(q = bc + ca + ab\) and \(r = abc\) as required.

  1. \begin{align*} \ln (1+qx^2 + rx^3) &= \ln(1+ax) + \ln(1+bx)+\ln(1+cx) \\ &= -\sum_{n=1}^{\infty} \frac{(-ax)^n}{n}-\sum_{n=1}^{\infty} \frac{(-bx)^n}{n}-\sum_{n=1}^{\infty} \frac{(-cx)^n}{n} \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(a^n+b^n+c^n)}{n} x^n \\ &= \sum_{n=1}^{\infty} (-1)^{n+1} S_n x^n \\ \end{align*}
  2. \begin{align*} \ln (1 + qx^2 + rx^3) &= (qx^2+rx^3) -\frac{(qx^2+rx^3)^2}{2} + O(x^6) \\ &= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + O(x^6) \\ \end{align*} Comparing coefficients we see that \(S_2 = -q\) and \(S_3 = r\), we also must have \(S_5 = -qr = S_2S_3\) as required.
  3. \begin{align*} \ln (1 + qx^2 + rx^3) &= (qx^2+rx^3) -\frac{(qx^2+rx^3)^2}{2} +\frac{(qx^2+rx^3)^3}{3}+ O(x^8) \\ &= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + \frac12 rx^6 + \frac13 q^3 x^6 + q^2r x^7 + O(x^8) \\ &= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + \left ( \frac12 r+ \frac13 q^3 \right)x^6 + q^2r x^7 \end{align*} Comparing coefficients we see that \(S_2 = -q\) and \(S_5 =-qr\), we also must have \(S_7 = q^2r = S_2S_5\) as required.
  4. Let \(a = b = 1, c = -2\), then \(S_2 = \frac{1^2+1^2 + (-2)^2}{2} = 3, S_7 = \frac{1^2+1^2+(-2)^7}{7} = -18, S_9 = \frac{1^1+1^2+(-2)^9}{9} = \frac{2 - 512}{9} \neq 3 \cdot (-18)\)

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2009 Paper 3 Q8
D: 1700.0 B: 1516.0
integration reduction formula integration by parts logarithmic limits series expansion exponential function

Let \(m\) be a positive integer and let \(n\) be a non-negative integer.

  1. Use the result \(\displaystyle \lim_{t\to\infty}\e^{-mt} t^n=0\) to show that \[ \lim_{x\to0} x^m (\ln x)^n =0\,. \] By writing \(x^x\) as \(\e^{x\ln x}\) show that \[ \lim _{x\to0} x^x=1\,. \]
  2. Let \(\displaystyle I_{n} = \int_0^1 x^m (\ln x)^n \d x\,\). Show that \[ I_{n+1} = - \frac {n+1}{m+1} I_{n} \] and hence evaluate \(I_{n}\).
  3. Show that \[ \int_0^1 x^x \d x = 1 -\left(\tfrac12\right)^2 +\left(\tfrac13\right)^3 -\left(\tfrac14\right)^4 + \cdots \,. \]

Solution:

  1. \(\,\) \begin{align*} && \lim_{x \to 0} x^m(\ln x)^n &= \lim_{t \to \infty} (e^{-t})^m (\ln e^{-t})^n \\ &&&= \lim_{t \to \infty} e^{-mt} (-t)^n \\ &&&= (-1)^n \lim_{t \to \infty} e^{-mt} t^n = 0 \\ \\ && \lim_{x \to 0} x^x &= \lim_{x \to 0} e^{x \ln x} \\ &&&= \exp \left (\lim_{x \to 0} x \ln x \right) \\ &&&= \exp \left (0 \right) = 1 \end{align*}
  2. \(\,\) \begin{align*} && I_{n} &= \int_0^1 x^m (\ln x)^n \d x \\ && I_{n+1} &= \int_0^1 x^m (\ln x)^{n+1} \d x \\ &&&= \left [\frac{x^{m+1}}{m+1} (\ln x)^{n+1} \right]_0^1 - \frac{1}{m+1} \int_0^1 x^{m+1} (n+1) (\ln x)^n \cdot x^{-1} \d x \\ &&&= 0 - \frac{1}{m+1} \lim_{x \to 0} \left (x^{m+1} (\ln x)^{n+1} \right) - \frac{n+1}{m+1} \int_0^1 x^m (\ln x)^n \d x \\ &&&= - \frac{n+1}{m+1} I_n \\ \\ && I_0 &= \int_0^1 x^m \d x = \frac{1}{m+1} \\ && I_1 &= -\frac{1}{(m+1)^2} \\ && I_2 &= \frac{2}{(m+1)^3} \\ && I_n &= (-1)^n \frac{n!}{(m+1)^{n+1}} \end{align*}
  3. \(\,\) \begin{align*} && \int_0^1 x^x \d x &= \int_0^1 e^{x \ln x} \d x \\ &&&= \int_0^1 \sum_{k=0}^{\infty} \frac{x^k(\ln x)^k}{k!} \d x \\ &&&= \sum_{k=0}^{\infty} \int_0^1 \frac{x^k(\ln x)^k}{k!} \d x\\ &&&= \sum_{k=0}^{\infty} (-1)^k \frac{k!}{k!(k+1)^{k+1}}\\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^{k+1}}\\ &&&= 1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + \cdots \end{align*}

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2008 Paper 2 Q2
D: 1600.0 B: 1498.5
partial fractions series expansion binomial expansion geometric series power series decimal approximation rational function

Let \(a_n\) be the coefficient of \(x^n\) in the series expansion, in ascending powers of \(x\), of \[\displaystyle \frac{1+x}{(1-x)^2(1+x^2)} \,, \] where \(\vert x \vert <1\,\). Show, using partial fractions, that either \(a_n =n+1\) or \(a_n = n+2\) according to the value of \(n\). Hence find a decimal approximation, to nine significant figures, for the fraction \( \displaystyle \frac{11\,000}{8181}\). \newline [You are not required to justify the accuracy of your approximation.]

Solution: \begin{align*} && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{Cx+D}{1+x^2} \\ \Rightarrow && 1 + x &= A(1-x)(1+x^2) + B(1+x^2) + Cx(1-x)^2 + D(1-x)^2 \\ \Rightarrow && 2 &= 2B \tag{\(x = 1\)} \\ \Rightarrow && 1 &= B \\ \Rightarrow && 1 &= A+B+D \tag{\(x = 0\)}\\ \Rightarrow && A &= -D \\ \Rightarrow && 0 &= 4A+2B-4C+4D \tag{\(x = -1\)}\\ \Rightarrow && C &= \frac12\\ \Rightarrow && 3 &= -5A+5B+2C+D \tag{\(x=2\)} \\ \Rightarrow && 3 &= -6A+6 \\ \Rightarrow && A,D &=-\frac12,\frac12 \\ \Rightarrow && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{1}{(1-x)^2} +\frac{1}{2(1-x)}+ \frac{x-1}{2(1+x^2)} \\ &&&=\sum_{k=0}^{\infty}(k+1)x^k + \sum_{k=0}^{\infty}\frac12 x^k + \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k+1} - \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k} \end{align*} Therefore the coefficient of \(x^n\) is \(n+1\) or \(n+2\) depending on whether the coefficients from the final series add constructively \(n \equiv 1, 2 \pmod{4}\) or destructively. \begin{align*} \frac{11\, 000}{8181} &= \frac{(10+1) \cdot 1000}{(10-1)^2 \cdot (100+1)} \\ &= \frac{(1+\frac{1}{10})}{(1-\frac{1}{10})^2 \cdot (1+\frac1{10})} \\ &= 1 + \frac3{10} + \frac4{10^2} + \frac{4}{10^3}+\frac{5}{10^4} + \frac{7}{10^5} + \frac{8}{10^6} + \cdots \\ & \quad \quad \cdots + \frac{8}{10^7} + \frac{9}{10^8} + \frac{11}{10^9} + \frac{12}{10^{10}} + \cdots \\ &= 1.34457890 + \frac{12}{10^{10}} + \cdots \end{align*} \begin{align*} && \sum_{k=m}^{\infty} (k+2)x^k &= x^m \sum_{k=0}^{\infty} (k+m+2)x^{k} \\ && &= \frac{x^k}{(1-x)^2} + \frac{(m+2)x^k}{1-x} \\ \Rightarrow && |\sum_{k=m}^{\infty} a_k \left ( \frac1{10} \right )^k | &\leq \frac{1}{10^m}\left ( \frac{1}{(1-\frac1{10})^2} + \frac{m+2}{1-\frac1{10}} \right) \\ &&&= \frac{1}{10^{m-1}} \left ( \frac{9m+28}{81}\right ) \end{align*} Therefore for this will be less than \(10^{-9}\), when \(m = 11\), so our approximation is valid to 9sf

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2007 Paper 2 Q1
D: 1600.0 B: 1516.0
generalised binomial theorem binomial expansion fractional exponent square root approximation cube root approximation numerical approximation series expansion

In this question, you are not required to justify the accuracy of the approximations.

  1. Write down the binomial expansion of \(\displaystyle \left( 1+\frac k {100} \right)^{\!\frac12}\)in ascending powers of \(k\), up to and including the \(k^3\) term.
    1. Use the value \(k=8\) to find an approximation to five decimal places for \(\sqrt{3}\,\).
    2. By choosing a suitable integer value of \(k\), find an approximation to five decimal places for \(\sqrt6\,\).
  2. By considering the first two terms of the binomial expansion of \(\displaystyle \left( 1+\frac k {1000} \right)^{\!\frac13}\), show that \(\dfrac{3029}{2100}\) is an approximation to \(\sqrt[3]{3}\).

Solution:

  1. Using the generalise binomial theorem \begin{align*} \left( 1+\frac k {100} \right)^{\frac12} &= 1 + \frac12 \frac{k}{100} + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)}{2!} \left (\frac{k}{100} \right)^2 + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)\cdot \left ( -\tfrac32\right)}{3!} \left (\frac{k}{100} \right)^3 + \cdots \\ &= 1 + \frac{1}{200}k - \frac{1}{80\,000}k^2 + \frac{1}{16\,000\,000}k^3 + \cdots \end{align*}
    1. If \(k = 8\), \begin{align*} && \left( 1+\frac 8 {100} \right)^{\frac12} &= 1 + \frac{1}{200}8 - \frac{1}{80\,000}8^2 + \frac{1}{16\,000\,000}8^3 + \cdots \\ \Rightarrow && \frac{6\sqrt{3}}{10} &\approx 1 + 0.04 - 0.0008 + 0.000032 \\ &&&= 1.039232\\ \Rightarrow && \sqrt{3} &\approx 1.73205 \, (5\, \text{d.p.}) \end{align*}
    2. If \(k = -4\), \begin{align*} && \left( 1-\frac 4 {100} \right)^{\frac12} &= 1 - \frac{1}{200}4 - \frac{1}{80\,000}4^2 - \frac{1}{16\,000\,000}4^3 + \cdots \\ \Rightarrow && \frac{4\sqrt{6}}{10} &\approx 1 -0.02-0.0002 -0.000004 \\ &&&= 0.979796\\ \Rightarrow && \sqrt{6} &\approx 2.44949\, (5\, \text{d.p.}) \end{align*}
  2. \(\,\) \begin{align*} && \left( 1+\frac k {1000} \right)^{\!\frac13} &= 1 + \frac13 \frac{k}{1000} + \cdots \\ &&&= 1 + \frac{k}{3\,000} + \cdots \\ && 3 \times 7^3 &= 1029 \\ \Rightarrow && \left( 1+\frac {29} {1000} \right)^{\!\frac13} &\approx 1 + \frac{29}{3\,000} \\ \Rightarrow && \frac{7\sqrt[3]{3}}{10} &\approx \frac{3\,029}{3000} \\ \Rightarrow && \sqrt[3]{3} &= \frac{3\,029}{2\,100} \end{align*}

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1997 Paper 3 Q1
D: 1700.0 B: 1500.0
Taylor series exponential function series expansion polynomial multiplication summation identity Maclaurin series

  1. By considering the series expansion of \((x^2+5x+4){\rm \; e}^x\) show that \[10{\rm\, e}=4+\frac{3^2}{1!}+\frac{4^2}{2!}+\frac{5^2}{3!}+\cdots\;.\]
  2. Show that \[5{\rm\, e}=1+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\cdots\;.\]
  3. Evaluate \[1+\frac{2^3}{1!}+\frac{3^3}{2!}+\frac{4^3}{3!}+\cdots\;.\]

Solution:

  1. \begin{align*} (x^2+5x+4)e^x &= \sum_{k=0}^\infty \frac{1}{k!} x^{k+2}+\sum_{k=0}^\infty \frac{5}{k!} x^{k+1}+\sum_{k=0}^\infty \frac{4}{k!} x^{k} \\ &= \sum_{k=0}^{\infty} \l \frac{1}{k!}+\frac{5}{(k+1)!}+\frac{4}{(k+2)!} \r x^{k+2} + 5x+4+4x \\ &= 4 + 9x + \sum_{k=0}^{\infty} \l \frac{(k+2)(k+1)}{(k+2)!}+\frac{5(k+2)}{(k+2)!}+\frac{4}{(k+2)!} \r x^{k+2}\\ &= 4 + 9x + \sum_{k=0}^{\infty} \l \frac{k^2+3k+2+5k+10+4}{(k+2)!} \r x^{k+2}\\ &= 4 + 9x + \sum_{k=0}^{\infty} \frac{(k+4)^2}{(k+2)!} x^{k+2}\\ &= 4 + 9x + \sum_{k=2}^{\infty} \frac{(k+2)^2}{k!} x^{k}\\ \end{align*} So when \(x = 1\) we have \[10e = 4 + \frac{3^2}{1!} + \frac{4^2}{2!} + \frac{5^2}{3!} + \cdots \]
  2. \begin{align*} (x^2+3x+1)e^x &= \sum_{k=0}^\infty \frac{1}{k!}x^{k+2}+\sum_{k=0}^\infty 3\frac{1}{k!}x^{k+1} + \sum_{k=0}^{\infty} \frac{1}{k!} x^k \\ &= 1+3x+\sum_{k=1}^{\infty} \l \frac1{(k-1)!}+\frac{3}{k!} + \frac{1}{(k+1)!} \r x^{k+1} \\ &= 1+3x+\sum_{k=1}^{\infty} \frac{(k+1)k + 3(k+1)+1}{(k+1)!}x^{k+1} \\ &= 1+3x+\sum_{k=1}^{\infty} \frac{k^2+4k+4}{(k+1)!}x^{k+1} \\ &= 1+3x+\sum_{k=0}^{\infty} \frac{(k+2)^2}{(k+1)!}x^{k+1} \\ &=1+3x+ \sum_{k=1}^{\infty} \frac{(k+1)^2}{k!}x^k \end{align*} Plugging in \(x=1\) we get the desired result.
  3. \begin{align*} && xe^x &= \sum_{k=0}^{\infty} \frac{x^{k+1}}{k!} \\ x\frac{\d}{\d x} : && x(1+x)e^x &= \sum_{k=0}^{\infty} \frac{(k+1)x^{k+1}}{k!} \\ x\frac{\d}{\d x} : && x(x(1+x)+1+2x)e^x &= \sum_{k=0}^{\infty} \frac{(k+1)^2x^{k+1}}{k!} \\ &&(x^3+3x^2+x)e^x &= \sum_{k=0}^{\infty} \frac{(k+1)^2x^{k+1}}{k!} \\ \frac{\d}{\d x} : && e^x(x^3+3x^2+x+3x^2+6x+1) &=\sum_{k=0}^{\infty} \frac{(k+1)^3x^{k}}{k!} \\ \Rightarrow && 15e &= 1 + \frac{2^3}{1!} + \frac{3^3}{2!} + \cdots \end{align*}

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