Year: 2024
Paper: 2
Question Number: 2
Course: LFM Stats And Pure
Section: Generalised Binomial Theorem
Many candidates produced good solutions to the questions, with the majority of candidates opting to focus on the pure questions of the paper. Candidates demonstrated very good ability, particularly in the area of manipulating algebra. Many candidates produced clear diagrams which in many cases meant that they were more successful in their attempts at their questions than those who did not do so. The paper also contained a number of places where the answer to be reached was given in the question. In such cases, candidates must be careful to ensure that they provide sufficient evidence of the method used to reach the result in order to gain full credit.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
In this question, you need not consider issues of convergence.
\begin{questionparts}
\item Find the binomial series expansion of $(8 + x^3)^{-1}$, valid for $|x| < 2$.
Hence show that, for each integer $m \geqslant 0$,
\[ \int_0^1 \frac{x^m}{8 + x^3}\,\mathrm{d}x = \sum_{k=0}^{\infty} \left( \frac{(-1)^k}{2^{3(k+1)}} \cdot \frac{1}{3k + m + 1} \right). \]
\item 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.
\item 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.
\end{questionparts}\begin{questionparts}
\item 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*}
\item 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*}
\item 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*}
\end{questionparts}
This question was attempted by approximately three-quarters of the candidates and many very good solutions were seen. In part (i) candidates were generally able to find the binomial series expansion, but some did not give a clear enough statement of the general term. A number of candidates did not recognize that it would be possible to use the series expansion to establish the integration result that was required and instead attempted to use integration by parts or to produce a proof by induction. Most candidates recognized the way in which part (i) could be applied to answer part (ii) and this part was generally answered well. A number of candidates forgot to evaluate the integral before moving on to the next part of the question. In part (iii) a large number of candidates recognized that the use of partial fractions would allow them to apply part (i) again, but many did not then realise that there was a need for partial fractions to be applied a second time. Almost all candidates who reached the final integral were able to recognize it and either state the answer or use an appropriate substitution.