Year: 2014
Paper: 2
Question Number: 8
Course: LFM Stats And Pure
Section: Binomial Theorem (positive integer n)
There were good solutions presented to all of the questions, although there was generally less success in those questions that required explanations of results or the use of diagrams and graphs to reach the solution. Algebraic manipulation was generally well done by many of the candidates although a range of common errors such as confusing differentiation and integration and simple arithmetic slips were evident. Candidates should also be advised to use the methods that are asked for in questions unless it is clear that other methods will be accepted (such as by the use of the phrase "or otherwise").
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1486.3
Banger Comparisons: 1
For positive integers $n$, $a$ and $b$, the integer $c_r$ ($0\le r\le n$) is defined to be the coefficient of $x^r$ in the expansion in powers of $x$ of $(a+bx)^n$. Write down an expression for $c_r$ in terms of $r$, $n$, $a$ and $b$. For given $n$, $a$ and $b$, let $m$ denote a value of $r$ for which $c_r$ is greatest (that is, $c_m \ge c_r$ for $0\le r\le n$).
Show that
\[
\frac{b(n+1)}{a+b} - 1 \le m \le \frac {b(n+1)}{a+b} \,.
\]
Deduce that $m$ is either a unique integer or one of
two consecutive integers. Let $G(n,a,b)$ denote the unique value of $m$ (if there is one) or the larger of the two possible values of $m$.
\begin{questionparts}
\item Evaluate $G(9,1,3)$ and $G(9,2,3)$.
\item For any positive integer $k$, find $G(2k,a,a)$ and $G(2k-1,a,a)$ in terms of $k$.
\item For fixed $n$ and $b$, determine a value of $a$ for which $G(n,a,b)$ is greatest.
\item For fixed $n$, find the greatest possible value of $G(n,1,b)$. For which values of $b$ is this greatest value achieved?
\end{questionparts}$c_r = \binom{n}{r}a^{n-r}b^r$
\begin{align*}
&& c_m &\geq c_{m+1} \\
\Rightarrow && \binom{n}{m} a^{n-m}b^m &\geq \binom{n}{m+1} a^{n-m-1}b^{m+1} \\
\Rightarrow && \frac{1}{(n-m)}a &\geq \frac{1}{m+1}b \\
\Rightarrow && (m+1)a &\geq (n-m)b \\
\Rightarrow && m(a+b) &\geq nb -a \\
\Rightarrow && m &\geq \frac{n(b+1)-a-b}{a+b}=\frac{n(b+1)}{a+b} - 1 \\
\\
&& c_m &\geq c_{m-1} \\
\Rightarrow && \binom{n}{m} a^{n-m}b^m &\geq \binom{n}{m-1} a^{n-m+1}b^{m-1} \\
\Rightarrow && \frac{1}m b &\geq \frac{a}{(n-m+1)} \\
\Rightarrow && (n-m+1)b &\geq ma \\
\Rightarrow && (n+1)b &\geq m(a+b) \\
\Rightarrow && m &\leq \frac{(n+1)b}{a+b}
\end{align*}
Since $m$ lies between two values $1$ apart is is either equal to one of those two values or is the unique integer between them.
Let $\displaystyle G(n,a,b) = \left \lfloor \frac{b(n+1)}{a+b} \right \rfloor$, so
\begin{questionparts}
\item $\,$ \begin{align*}
&& G(9,1,3) &= \left \lfloor \frac{3(9+1)}{1+3} \right \rfloor \\
&&&= \left \lfloor \frac{30}{4} \right \rfloor \\
&&&= 7 \\
\\
&& G(9,2,3) &= \left \lfloor \frac{3(9+1)}{2+3} \right \rfloor \\
&&&= \left \lfloor \frac{30}{5} \right \rfloor \\
&&&= 6
\end{align*}
\item $\,$ \begin{align*}
&& G(2k, a, a) &= \left \lfloor \frac{a(2k+1)}{a+a} \right \rfloor \\
&& &= \left \lfloor \frac{2k+1}{2} \right \rfloor \\
&&&= k \\
\\
&& G(2k-1, a, a) &= \left \lfloor \frac{a(2k-1+1)}{a+a} \right \rfloor \\
&& &= \left \lfloor k\right \rfloor \\
&&&= k \\
\end{align*}
\item $G(n,a,b)$ is decreasing in $a$, therefore take $a = 1$.
\item For fixed $n$, we are looking at $\frac{a(n+1)}{a+b}$ and we want this to be as large as possible. By considering $(n+1) - \frac{b(n+1)}{a+b}$ we can see this is increasing in $b$ and the largest value possible is $n$. This is achieved when
\begin{align*}
&& \frac{b(n+1)}{a+b} & \geq n \\
\Leftrightarrow && bn + b &\geq na + bn \\
\Leftrightarrow && b& \geq na
\end{align*}
\end{questionparts}
This was the least popular of the pure maths questions and also the one with the lowest average score. Many of the candidates were able to show the required result at the start of the question, although very few candidates explained that n could be either of the two integers when the range included two integers. Parts (i) and (ii) were then quite straightforward for most candidates, although many calculated the range of values but did not justify their choice in the case where there were two possibilities. In the final two parts of the question some candidates mistakenly chose the value 0 when asked for a positive integer.