Year: 2010
Paper: 1
Question Number: 5
Course: LFM Stats And Pure
Section: Binomial Theorem (positive integer n)
There were significantly more candidates attempting this paper than last year (just over 1000), and the scores were much higher than last year (presumably due to the easier first question): fewer than 2% of candidates scored less than 20 marks overall, and the median mark was 61. The pure questions were the most popular as usual, though there was much more variation than in some previous years: questions 1, 3, 4 and 6 were the most popular, while question 7 (on vectors) was intensely unpopular. About half of all candidates attempted at least one mechanics question, and 15% attempted at least one probability question. The marks were unsurprising: the pure questions generally gained the better marks, while the mechanics and probability questions generally had poorer marks. A sizeable number of candidates ignored the advice on the front cover and attempted more than six questions, with a fifth of candidates trying eight or more questions. A good number of those extra attempts were little more than failed starts, but suggest that some candidates are not very effective at question-picking. This is an important skill to develop during STEP preparation. Nevertheless, the good marks and the paucity of candidates who attempted the questions in numerical order does suggest that the majority are being wise in their choices. Because of the abortive starts, I have often restricted my attention below to those attempts which counted as one of the six highest-scoring answers, and referred to these as "significant attempts". The majority of candidates did begin with question 1 (presumably as it appeared to be the easiest), but some spent far longer on it than was wise. Some attempts ran to over eight pages in length, especially when they had made an algebraic slip early on, and used time which could have been far better spent tackling another question. It is important to balance the desire to finish the question with an appreciation of when to stop and move on. Many candidates realised that for some questions, it was possible to attempt a later part without a complete (or any) solution to an earlier part. An awareness of this could have helped some of the weaker students to gain vital marks when they were stuck; it is generally better to do more of one question than to start another question, in particular if one has already attempted six questions. It is also fine to write "continued later" at the end of a partial attempt and then to continue the answer later in the answer booklet. As usual, though, some candidates ignored explicit instructions to use the previous work, such as "Hence", or "Deduce". They will get no credit if they do not do what they are asked to! (Of course, "Hence, or otherwise, show . . ." gives them the freedom to use any method of their choosing; often the "hence" will be the easiest, but in Question 5 this year, the "otherwise" approach was very popular.) On some questions, some candidates tried to work forwards from the given question and backwards from the answer, hoping that they would meet somewhere in the middle. While this worked on occasion, it often required fudging. It is wise to remember that STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Difficulty Rating: 1484.0
Difficulty Comparisons: 1
Banger Rating: 1484.0
Banger Comparisons: 1
By considering the expansion
of $\left(1+x\right)^{n}$ where $n$ is a positive integer,
or otherwise, show that:
\begin{questionparts}
\item \[\binom{n}{0}+\binom{n}1+\binom{n}2 +\cdots +\binom{n}n=2^{n} \]
\item \[\binom{n}{1}+2\binom{n}2+3\binom{n}3 +\cdots +n\binom{n}n=n2^{n-1} \]
\item \[\binom{n}{0}+\frac12\binom{n}1+\frac13\binom{n}2 +\cdots +\frac1{n+1}\binom{n}n=\frac1{n+1}(2^{n+1}-1) \]
\item \[\binom{n}{1}+2^2\binom{n}2+3^2\binom{n}3 +\cdots +n^2\binom{n}n=n(n+1)2^{n-2} \]
\end{questionparts}\begin{questionparts}
\item Notice that
\begin{align*}
&& (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\
\text{Evaluate at }x = 1: && 2^n &= \sum_{i=0}^n \binom{n}{i}
\end{align*}
\item $\,$
\begin{align*}
&& (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\
\frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{i=1}^n i\binom{n}{i} x^{i-1} \\
\text{Evaluate at }x = 1: && n2^{n-1} &= \sum_{i=1}^n i\binom{n}{i}
\end{align*}
\item $\,$
\begin{align*}
&& (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\
\Rightarrow && \int_0^1(1+x)^n \d x &= \int_0^1 \sum_{i=0}^n \binom{n}{i} x^i \d x \\
\Rightarrow && \frac{1}{n+1}(2^{n+1}-1) &= \sum_{i=0}^n \binom{n}{i}\int_0^1 x^i \d x\\
&&& = \sum_{i=0}^n \frac{1}{i+1}\binom{n}{i} \\
\end{align*}
\item $\,$
\begin{align*}
&& (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\
\frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{i=1}^n i\binom{n}{i} x^{i-1} \\
\times x: && nx(1+x)^{n-1} &= \sum_{i=1}^n i\binom{n}{i} x^{i} \\
\frac{\d}{\d x}: && n(1+x)^{n-1}+n(n-1)x(1+x)^{n-2} &= \sum_{i=1}^n i^2\binom{n}{i} x^{i-1} \\
\text{Evaluate at }x = 1: && \sum_{i=1}^n i^2\binom{n}{i} &= n(1+1)^{n-1}+n(n-1)x(1+1)^{n-2} \\
&&&= 2^{n-2} \left (n(n-1) + 2n \right) \\
&&&= n(n+1)2^{n-2}
\end{align*}
\end{questionparts}
This was a fairly unpopular question, attempted by only about 40% of the candidates (and being one of the best six attempts of fewer than one-third of candidates). The marks were also poor, with the median mark of the significant attempts being 3/20. About one-sixth of all attempts failed to gain a single mark, even though writing out the binomial expansion of (1 + x)n using binomial coefficients would have been sufficient for this. About one-quarter of candidates failed to make any further progress beyond this point, though another quarter substituted x = 1 and reached the result of part (i) before giving up. Those who persevered generally did reasonably well, with many correct answers to parts (ii) and (iii), and a good number succeeding on part (iv) as well. It was surprising how many successfully answered parts (ii) and (iii) using binomial coefficient manipulations without any reference back to part (i). In part (ii), a number of candidates attempted to differentiate 2n to get n.2n−1. Others, who used the binomial manipulation approach, were careless about the first and last terms of the sum or made no attempt to justify why all of the corresponding middle terms were equal. On this note, some candidates expanded both sides of the identity they were trying to prove and equated the corresponding terms without any understanding or indication of why they were the same; this gained relatively few marks, especially in part (iv) where the terms do not even correspond in this way. In part (iii), those candidates who used a calculus-based approach frequently failed to consider the constant of integration, and could not, therefore, justify the need for the −1 term. Finally, whereas a good number of the calculus-based approaches succeeded on part (iv), no candidate who used a binomial-manipulation method managed to extend their techniques to this case.