Year: 2016
Paper: 2
Question Number: 5
Course: LFM Stats And Pure
Section: Generalised Binomial Theorem
As in previous years the Pure questions were the most popular of the paper with questions 1, 3 and 7 the most popular of these. The least popular questions of the paper were questions 10, 11, 12 and 13 with fewer than 400 attempts for each of them. There were many examples of solutions in this paper that were insufficiently well explained, given that the answer to be reached had been provided in the question.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
In this question, the definition of $\displaystyle\binom pq$
is taken to be
\[
\binom pq =
\begin{cases}
\dfrac{p!}{q!(p-q)!} & \text{ if } p\ge q\ge0 \,,\\[4mm]
0 & \text{ otherwise } .
\end{cases}
\]
\begin{questionparts}
\item
Write down the coefficient of $x^n$ in the binomial expansion
for $(1-x)^{-N}$, where $N$ is a positive integer, and write
down the expansion
using the $\Sigma$ summation notation.
By considering
$
(1-x)^{-1} (1-x)^{-N}
\,
,$
where $N$ is a positive integer, show that
\[
\sum_{j=0}^n \binom { N+j -1}{j} = \binom{N+n}{n}\,.
\]
\item
Show that,
for any positive integers $m$, $n$ and $r$ with $r\le m+n$,
\[
\binom{m+n} r = \sum _{j=0}^r \binom m j \binom n {r-j}
\,.
\]
\item
Show that, for any positive integers $m$ and $N$,
\[
\sum_{j=0}^n(-1)^{j} \binom {N+m} {n-j} \binom {m+j-1}{j } =
\displaystyle \binom N n
.
\]
\end{questionparts}\begin{questionparts}
\item $\frac{(-N)(-N-1)\cdots(-N-n+1)}{n!} = \binom{N+n-1}{n}$, so
\[ (1-x)^{-N} = \sum_{n=0}^{\infty} \binom{N+n-1}{n} x^n\]
\begin{align*}
&& (1-x)^{-N-1} &= (1-x)^{-1}(1-x)^{-N} \\
&&&= (1 + x + x^2 + \cdots)\left ( \sum_{n=0}^{\infty} \binom{N+n-1}{n} x^n\right)\\
[x^{n}]: && \binom{N+1+n-1}{n} &= \sum_{j=0}^n \underbrace{1}_{x^{n-j} \text{ from 1st bracket}}\cdot\underbrace{\binom{N+j-1}{j}}_{x^j\text{ from second bracket}} \\
\Rightarrow && \binom{N+n}{n} &= \sum_{j=0}^n \binom{N+j-1}{j}
\end{align*}
\item Consider $(1+x)^{m+n} = (1+x)^m(1+x)^n$ and consider the coefficient of $x^r$ from each side. On the left hand side this is clearly $\binom{m+n}{r}$ on the right hand side we can take $x^j$ from $(1+x)^m$ and $x^{n-j}$ from $(1+x)^n$ and $j$ can take any value from $0$ to $r$, ie
\[ \binom{m+n} r = \sum _{j=0}^r \binom m j \binom n {r-j} \]
\item Consider $(1-x)^{-(N+m+1)} = ($
\end{questionparts}
This was the least attempted of all of the Pure questions and one that generally produced low marks for candidates, who generally appeared to have difficulty in expressing the coefficients of the binomial expansion in the required form. In each case the desired answer was given in the question, and so successful solutions also needed to be very clear about the reasoning used to reach the answer.