Year: 2024
Paper: 2
Question Number: 6
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 The sequence $T_n$, for $n = 0, 1, 2, \ldots$, is defined by $T_0 = 1$ and, for $n \geqslant 1$, by
\[ T_n = \frac{2n-1}{2n}\,T_{n-1}. \]
Prove by induction that
\[ T_n = \frac{1}{2^{2n}}\binom{2n}{n}, \]
for $n = 0, 1, 2, \ldots$.
[Note that $\dbinom{0}{0} = 1$.]
\item Show that in the binomial series for $(1-x)^{-\frac{1}{2}}$,
\[ (1-x)^{-\frac{1}{2}} = \sum_{r=0}^{\infty} a_r x^r, \]
successive coefficients are related by
\[ a_r = \frac{2r-1}{2r}\,a_{r-1} \]
for $r = 1, 2, \ldots$\,.
Hence prove that $a_r = T_r$ for all $r = 0, 1, 2, \ldots$\,.
\item Let $b_r$ be the coefficient of $x^r$ in the binomial series for $(1-x)^{-\frac{3}{2}}$, so that
\[ (1-x)^{-\frac{3}{2}} = \sum_{r=0}^{\infty} b_r x^r. \]
By considering $\dfrac{b_r}{a_r}$, find an expression involving a binomial coefficient for $b_r$, for $r = 0, 1, 2, \ldots$\,.
\item By considering the product of the binomial series for $(1-x)^{-\frac{1}{2}}$ and $(1-x)^{-1}$, prove that
\[ \frac{(2n+1)}{2^{2n}}\binom{2n}{n} = \sum_{r=0}^{n} \frac{1}{2^{2r}}\binom{2r}{r}, \]
for $n = 1, 2, \ldots$\,.
\end{questionparts}\begin{questionparts}
\item Claim: $\displaystyle T_n = \frac{1}{2^{2n}}\binom{2n}{n}$
Proof: (By Induction)
Base case: $n=0$. Note that $T_0 = 1$ and $\frac{1}{2^0}\binom{0}{0} = 1$ so the base case is true.
Assume true for some $n=k$, ie $T_k = \frac{1}{2^{2k}} \binom{2k}{k}$ so
\begin{align*}
&& T_{k+1} &= \frac{2(k+1)-1}{2(k+1)} \frac{1}{2^{2k}} \binom{2k}{k} \\
&&&= \frac{2k+1}{k+1} \frac{1}{2^{2k+1}} \frac{(2k)!}{k!k!} \\
&&&= \frac{2(k+1)(2k+1)}{(k+1)(k+1)} \frac{1}{2^{2(k+1)}} \frac{(2k)!}{k!k!} \\
&&&= \frac{1}{2^{2(k+1)}} \frac{(2k+2)!}{(k+1)!(k+1)!} \\
&&&= \frac{1}{2^{2(k+1)}} \binom{2(k+1)}{k+1}
\end{align*}
and therefore it's true for all $n$.
\item Notice that $(1-x)^{-\frac12} = 1 + (-\tfrac12)(-x) + \frac{(-\frac12)(-\frac32)}{2!}(-x)^2+\cdots$ in particular $a_r = \frac{(-\frac12 - r)}{r}(-1)a_{r-1} = \frac{2r-1}{2r}a_{r-1}$. Since $a_0 = 1$ we have $a_r = T_r$ for all $r$.
\item Notice that \begin{align*}
&& (1-x)^{-\frac32} &= \sum_{r=0}^\infty b_r x^r \\
&&&= \sum_{r=0}^\infty \frac{(-\frac32)\cdot(-\frac32-1)\cdots (-\frac32-(r-1))}{r!}(-x)^r \\
&&&= \sum_{r=0}^\infty \frac{(-\frac12-1)\cdot(-\frac12-2)\cdots (-\frac12-r)}{r!}(-x)^r \\
\end{align*}
Therefore $\frac{b_r}{a_r} = \frac{r+\frac12}{\frac12} = 2r+1$ so $b_r = \frac{2r+1}{2^{2r}} \binom{2r}{r}$
\item Notice that \begin{align*}
&& (1-x)^{-\frac32} &= (1-x)^{-\frac12}(1-x)^{-1} \\
&&&= (1 + x+ x^2 + \cdots) \sum_{r=0}^{\infty} a_r x^r \\
&&&= \sum_{i=0}^{\infty} \sum_{k=0}^n a_r x^i
\end{align*}
So we must have $b_r = \sum_{i=0}^ra_i$ which is the required result
\end{questionparts}
A large number of attempts at this question were seen and many candidates were able to produce good solutions. Part (i) was answered well, although a number of candidates did not select the correct base case. Several solutions did not give sufficient detail in the proof to gain full credit however. Since the required form is known, it is important that steps in the solution are shown clearly. In part (ii) many candidates started by calculating some of the terms and then attempted to spot a pattern, or match the terms to the desired result. To gain full credit a solution that showed the form of the general term of the binomial series expansion was required. In a small number of cases combinatorial coefficients with non-integer arguments were used, but no explanation was given of the meaning of this notation. In part (iii) a number of candidates again did not write down the binomial coefficients and used pattern spotting. In general, candidates who had successfully answered part (ii) were also successful in part (iii). Some good solutions to part (iv) were seen, but often there was insufficient detail in the solutions to gain full credit.