Year: 2024
Paper: 2
Question Number: 1
Course: LFM Pure
Section: Proof
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 the equality
\[ 4 + 5 + 6 + 7 + 8 = 9 + 10 + 11, \]
the sum of the five consecutive integers from 4 upwards is equal to the sum of the next three consecutive integers.
Throughout this question, the variables $n$, $k$ and $c$ represent positive integers.
\begin{questionparts}
\item Show that the sum of the $n + k$ consecutive integers from $c$ upwards is equal to the sum of the next $n$ consecutive integers if and only if
\[ 2n^2 + k = 2ck + k^2. \]
\item Find the set of possible values of $n$, and the corresponding values of $c$, in each of the cases
\begin{enumerate}
\item $k = 1$
\item $k = 2$.
\end{enumerate}
\item Show that there are no solutions for $c$ and $n$ if $k = 4$.
\item Consider now the case where $c = 1$.
\begin{enumerate}
\item Find two possible values of $k$ and the corresponding values of $n$.
\item Show, given a possible value $N$ of $n$, and the corresponding value $K$ of $k$, that
\[ N' = 3N + 2K + 1 \]
will also be a possible value of $n$, with
\[ K' = 4N + 3K + 1 \]
as the corresponding value of $k$.
\item Find two further possible values of $k$ and the corresponding values of $n$.
\end{enumerate}
\end{questionparts}\begin{questionparts}
\item Suppose the sum of the $n + k$ consecutive integers from $c$ upwards is equal to the sum of the next $n$ consecutive integers then
\begin{align*}
&& \sum_{i=c}^{i=c+n+k-1} i &= \sum_{i=c+n+k}^{c+2n+k-1} i \\
\Leftrightarrow && \frac{(c+n+k-1)(c+n+k)}{2} - \frac{(c-1)c}{2} &= \frac{(c+2n+k-1)(c+2n+k)}{2} - \frac{(c+n+k-1)(c+n+k)}{2} \\
\Leftrightarrow && 2(c+n+k-1)(c+n+k) &= (c+2n+k-1)(c+2n+k) + c(c-1) \\
\Leftrightarrow && 2c^2+4cn+4ck+2n^2+4kn+2k^2-2c-2n-2k&=2c^2+4cn+2ck+4n^2+4nk+k^2-2c-2n-k \\
\Leftrightarrow && 2ck+k^2&=2n^2+k \\
\end{align*}
\item \begin{enumerate}
\item If $k=1$ then \begin{align*}
&& 2n^2 + 1 &= 2c + 1 \\
\Rightarrow && c &= n^2
\end{align*} So $n$ can take any value and $c = n^2$
\item If $k=2$ then \begin{align*}
&& 2n^2+2&= 4c+4 \\
\Rightarrow && n^2-1 &=2c
\end{align*} So $n$ must be odd, and $c = \frac12(n^2-1)$
\end{enumerate}
\item Suppose $k=4$ then $2n^2+4 = 8c+16$ or $n^2-6 = 4c$ but then the left hand side is $2, 3 \pmod{4}$ which is a contradiction.
\item Suppose $c =1$ \begin{enumerate}
\item Since $2n^2+k = 2k + k^2$ or $2n^2 = k^2+k$ we can have $k = 1, n = 1$ or $k = 8, n = 6$
\item Suppose $2N^2 = K^2 + K$ then consider
\begin{align*}
&& 2(N')^2 &= 2(3N+2K+1)^2 \\
&&&= 2(9N^2+4K^2+1+12NK+6N+4K) \\
&&&= 18N^2+8K^2+24NK+12N+8K+2 \\
&& (K')^2+K' &= (4N+3K+1)^2 + (4N+3K+1) \\
&&&= 16N^2 + 9K^2+1+24NK+12N+9K+1 \\
&&&= 16N^2+9K^2+24NK+12N+9K+2 \\
\Rightarrow && 2(N')^2-(K')^2-K' &= 2N^2-K^2-K \\
&&&= 0
\end{align*} as required.
\item So consider $(k,n) = (1,1), (8,6), (49, 35), (288,204)$
\end{enumerate}
\end{questionparts}
There were a large number of attempts at this question, with many good answers seen and many attempting most parts of the question. Many were able to show that the required formula in part (i) will be satisfied if the stated sum is satisfied, but many did not explain that the result applies in both directions with sufficient detail. Of those who failed to show the result in this part the main error was an incorrect choice of limits when expressing the sums either in sigma notation or as an arithmetic series. Candidates generally demonstrated an understanding of what was required for part (ii), but a significant number did not express their solution in a clear form, for example by finding just one case or listing the first few without specifying a general relationship. A number of candidates got confused between squares and square roots and having deduced that the square root of n is equal to c then concluded that n must be a square number. In part (iii) most candidates successfully identified the equation that needed to be satisfied, but were unable to explain clearly why there were no solutions. Part (iv)(a) was generally completed well, including by candidates who had struggled in earlier parts of the question. Many good answers to part (iv)(b) were also seen, although a number of cases did not explicitly identify the relationship that must exist between the values of N and K. Most candidates recognised that the result of part (iv)(b) could be used to generate the further solutions required in part (iv)(c).