Year: 2017
Paper: 3
Question Number: 8
Course: UFM Pure
Section: Sequences and series, recurrence and convergence
The total entry was only very slightly smaller than that of 2016, which was a record entry, but was still over 10% more than 2015. No question was attempted by in excess of 90%, although two were very popular and also five others were attempted by 60% or more. No question was generally avoided with even the least popular one attracting more than 10% of the entry. Less than 10% of candidates attempted more than 7 questions, and, apart from 18 exceptions, those doing so did not achieve very good totals and seemed to be 'casting around' to find things they could do: the 18 exceptions were very strong candidates who were generally achieving close to full marks on all the questions they attempted. The general trend was that those with six attempts fared better than those with more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Prove that, for any numbers $a_1, a_2, \ldots\,,$
and $b_1, b_2, \ldots\,,$ and for $n\ge1$,
\[
\sum_{m=1}^n a_m(b_{m+1} -b_m) = a_{n+1}b_{n+1} -a_1b_1
-\sum_{m=1}^n b_{m+1}(a_{m+1} -a_m)
\,.
\]
\begin{questionparts}
\item
By setting $b_m = \sin mx$, show that
\[
\sum_{m=1}^n \cos (m+\tfrac12)x
= \tfrac12
\big(\sin (n+1)x - \sin x \big)
\cosec \tfrac12 x
\,.
\]
\textbf{Note:} $\sin A - \sin B = \displaystyle 2
\cos \big( \tfrac{{\displaystyle A+B\vphantom{_1}}}
{\displaystyle 2\vphantom{^1}} \big)\,
\sin\big( \tfrac{{\displaystyle A-B\vphantom{_1}}}{\displaystyle 2\vphantom{^1}} \big)\,
$.
\item
Show that
\[
\sum_{m=1}^n m\sin mx
=
\big (p \sin(n+1)x +q \sin nx\big)
\cosec^2 \tfrac12 x
\,,
\]
where $p$ and $q$ are to be determined in terms of $n$.
\textbf{Note:} $2\sin A \sin B = \cos (A-B) - \cos (A+B)\,$;
\textbf{Note:} $2\cos A \sin B = \sin (A+B) - \sin (A-B)\,$.
\end{questionparts}\begin{align*}
\sum_{m=1}^n a_m(b_{m+1} -b_m) +\sum_{m=1}^n b_{m+1}(a_{m+1} -a_m) &= \sum_{m=1}^n \left (a_{m+1}b_{m+1}-a_mb_m \right) \\
&= a_{n+1}b_{n+1} - a_1b_1
\end{align*}
And the result follows.
\begin{questionparts}
\item Let $b_m = \sin m x $, $a_m = \cosec \frac{x}{2}$, so
\begin{align*}
&& \sum_{m=1}^n \cosec \frac{x}{2} \left (\sin (m+1)x - \sin mx \right) &= \sum_{m=1}^n \cosec \frac{x}{2} 2 \cos \left ( \frac{2m+1}{2}x \right) \sin \left ( \frac{(m+1)-m}{2}x \right) \\
&&&=2 \sum_{m=1}^n\cos \left ( (m + \tfrac12)x \right)\\
\\
\Rightarrow && \sum_{m=1}^n\cos \left ( (m + \tfrac12)x \right) &= \tfrac12 \cosec \tfrac{x}{2}\left ( \sin(n+1)x - \sin x \right)
\end{align*}
\item $\,$ \begin{align*}
&& b_{m+1}-b_m &= \sin m x \sin \tfrac12 x \\
&&&= \frac12 \left ( \cos (m-\tfrac12)x - \cos (m+\tfrac12)x \right)\\
\Rightarrow && b_m &= -\tfrac12 \cos (m - \tfrac12)x\\
&& a_m &= m \\
\Rightarrow && \sum_{m=1}^n m \sin m x \sin \tfrac12 x &= (n+1) b_{n+1} - 1 \cdot b_1 - \sum_{m=1}^n b_{m+1} \cdot 1 \\
&&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+\tfrac12\cos(\tfrac12x) + \tfrac12\sum_{m=1}^n \cos(m+\tfrac12)x \\
&&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+\tfrac12\cos(\tfrac12x) + \tfrac14 \cosec \tfrac{x}{2}\left ( \sin(n+1)x - \sin x \right) \\
&&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+ \tfrac14 \cosec \tfrac{x}{2}\sin(n+1)x \\
&&&= \tfrac12\cosec\tfrac{x}2 \left (\tfrac12 \sin (n+1)x-(n+1)\cos(n+\tfrac12)x\sin\tfrac12x \right) \\
&&&= \tfrac12\cosec\tfrac{x}2 \left (\tfrac12 \sin (n+1)x-(n+1)\tfrac12 \left ( \sin (n+1)x - \sin nx \right) \right) \\
&&&= \tfrac14 \cosec \tfrac{x}{2} \left ( -n \sin (n+1)x +(n+1) \sin n x \right)
\end{align*}
Therefore $p = -\frac{n}4, q = \frac{n+1}{4}$
\end{questionparts}
Notice the connection here to integration by parts.
This was attempted as many times as question 5, but the success rate was about halfway between that of questions 5 and 7. Many attempts were made using induction which wasted a lot of time. Otherwise, in general, the stem and part (i) were well solved, but many could not spot the method to proceed with part (ii).