Year: 2016
Paper: 3
Question Number: 5
Course: LFM Stats And Pure
Section: Binomial Theorem (positive integer n)
A substantially larger number of candidates took the paper this year: 14% more than in 2015. However, the mean score was virtually identical to that in 2015. Five questions were very popular, with two being attempted by in excess of 90% of the candidates, but once again, all questions were attempted by significant numbers, with only one dipping under 10% attempting it, and every question was answered perfectly by at least one candidate. Most candidates kept to six sensible attempts, although some did several more scoring weakly overall, except in six outstanding cases that earned very high marks.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item By considering the binomial expansion of $(1+x)^{2m+1}$, prove that
\[
\binom{ 2m \! +\! 1}{ m} < 2^{2m}\,,
\]
for any positive integer $m$.
\item
For any positive integers $r$ and $s$ with $r< s$, $P_{r,s}$ is defined as follows: $P_{r,s}$ is the product of all the prime numbers greater than $r$ and less than or equal to $s$, if there are any such primes numbers; if there are no such primes numbers, then $P_{r,s}=1\,$.
For example, $P_{3,7}=35$, $P_{7,10}=1$ and $P_{14,18}=17$.
Show that, for any positive integer $m$,
$P_{m+1\,,\, 2m+1} $ divides $\displaystyle \binom{ 2m \! +\! 1}{ m} \,,$ and deduce that
\[
P_{m+1\,,\, 2m+1} < 2^{2m}
\,.
\]
\item Show that, if $P_{1,k} < 4^k$ for $k = 2$, $3$, $\ldots$, $2m$,
then $ P_{1,2m+1} < 4^{2m+1}\,$.
\item Prove that $\P_{1,n} < 4^n$ for $n\ge2$.
\end{questionparts}\begin{questionparts}
\item Notice that $(1+x)^{2m+1} = 1+\binom{2m+1}{1}x+\cdots + \binom{2m+1}{m}x^{m} + \binom{2m+1}{m+1} + \cdots$. Notice also that $\binom{2m+1}{m} = \binom{2m+1}{m+1}$. Therefore evaluating at $x = 1$, we see $2^{2m+1} > \binom{2m+1}{m} + \binom{2m+1}{m+1} = 2 \binom{2m+1}{m} \Rightarrow \binom{2m+1}{m} < 2^{2m}$
\item Each prime dividing $P_{m+1, 2m+1}$ divides the numerator of $\binom{2m+1}{m}$ since it appears in $(2m+1)!$, but not the denominator, since they wont appear in $m!$ or $(m+1)!$, and since they are prime they have to appear to divide it. Therefore the must divide $\binom{2m+1}{m}$ and therefore $P_{m+1,2m+1}$ must divide that binomail coefficient. Since $a \mid b \Rightarrow a \leq b$ we must have $P_{m+1, 2m+1} \leq \binom{2m+1}{m} < 2^{2m}$
\item Since
\begin{align*}
P_{1,2m+1} &= P_{1,m+1}P_{m+1, 2m+1} \tag{split into primes below $m+1$ and abvoe} \\
&< 4^{m+1}P_{m+1,2m+1} \tag{use the condition from the question}\\
&<4^{m+1}2^{2m} \tag{use our inequality} \\
&= 4^{2m+1}
\end{align*}
\item We proceed by (strong) induction.
Base case: ($n = 2$): $P_{1,2} = 2 < 4^2 =16$
Suppose it is true for all $k=2,3,\cdots,2m$ then it is true for $k=2m+1$ by the previous part of the question. However it is also true for $k=2m+2$, since that can never be prime (as n is now an even number bigger than 2).
Therefore by the principle of mathematical induction it is true for all $n$.
\end{questionparts}
This was attempted by half the candidature, scoring similarly to question 2. The binomial expansion and its symmetry were well-handled in the first part, as was applying the result of (iii) in part (iv). Part (ii) was not well-answered as there tended to be cavalier statements regarding divisibility, and in part (iii), few noticed that the condition p + 1 ≤ 2 was required in order to use (ii).