Year: 2025
Paper: 2
Question Number: 8
Course: LFM Pure
Section: Proof
As is commonly the case, the vast majority of candidates focused on the Pure questions in Section A of the paper, with a good number of attempts made on all of those questions. Candidates that attempted the Mechanics questions in Section B generally answered both questions. More candidates attempted Question 11 in Section C than either Mechanics question, but very few attempted Question 12 in that section. There were a large number of good responses seen for all the questions, but a significant number of responses lacked sufficient detail in the presentation, particularly when asked to prove a given result or provide an explanation. Candidates who did well on this paper generally: gave careful explanations of each step within their solutions; indicated all points of interest on graphs and other diagrams clearly; made clear comments about the approach that needed to be taken, particularly when having to explore a number of cases as part of the solution to a question; used mathematical terminology accurately within their solutions. Candidates who did less well on this paper generally: made errors with basic algebraic manipulation, such as incorrect processing of indices; produced sketches of graphs in which significant points were difficult to see clearly because of the chosen scale; skipped important lines within lengthy sections of algebraic reasoning.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
If we split a set $S$ of integers into two subsets $A$ and $B$ whose intersection is empty and whose union is the whole of $S$, and such that
\begin{itemize}
\item the sum of the elements of $A$ is equal to the sum of the elements of $B$
\item and the sum of the squares of the elements of $A$ is equal to the sum of the squares of the elements of $B$,
\end{itemize}
then we say that we have found a balanced partition of $S$ into two subsets.
\begin{questionparts}
\item Find a balanced partition of the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ into two subsets $A$ and $B$, each of size 4.
\item Given that $a_1, a_2, \ldots, a_m$ and $b_1, b_2, \ldots, b_m$ are sequences with
\[\sum_{k=1}^m a_k = \sum_{k=1}^m b_k \quad \text{and} \quad \sum_{k=1}^m a_k^2 = \sum_{k=1}^m b_k^2,\]
show that
\[\sum_{k=1}^m a_k^3 + \sum_{k=1}^m (c + b_k)^3 = \sum_{k=1}^m b_k^3 + \sum_{k=1}^m (c + a_k)^3\]
for any real number $c$.
\item Find, with justification, a balanced partition of the set $\{1, 2, 3, \ldots, 16\}$ into two subsets $A$ and $B$, each of size 8, which also has the property that
\begin{itemize}
\item the sum of the cubes of the elements of $A$ is equal to the sum of the cubes of the elements of $B$.
\end{itemize}
\item You are given that the sets $A = \{1, 3, 4, 5, 9, 11\}$ and $B = \{2, 6, 7, 8, 10\}$ form a balanced partition of the set $\{1, 2, 3, \ldots, 11\}$.
Let $S = \{n^2, (n+1)^2, (n+2)^2, \ldots, (n+11)^2\}$, where $n$ is any positive integer. Find, with justification, two subsets $C$ and $D$ of $S$ whose intersection is empty and whose union is the whole of $S$, and such that
\begin{itemize}
\item the sum of the elements of $C$ is equal to the sum of the elements of $D$.
\end{itemize}
\end{questionparts}\begin{questionparts}
\item $A = \{1,4,6,7 \}, B = \{2,3,5,8 \}$
\item \begin{align*}
&& \sum_{k=1}^m a_k^3 + \sum_{k=1}^m (c + b_k)^3 &= \sum_{k=1}^m a_k^3 + mc^3 + 3c^2 \sum_{k=1}^m b_k + 3c \sum_{k=1}^m b_k^2 + \sum_{k=1}^m b_k^3 \\
&&&= \sum_{k=1}^m a_k^3 + mc^3 + 3c^2 \sum_{k=1}^m a_k + 3c \sum_{k=1}^m a_k^2 + \sum_{k=1}^m b_k^3 \\
&&&= \sum_{k=1}^m a_k^3 + 3c \sum_{k=1}^m a_k^2+ 3c^2 \sum_{k=1}^m a_k + mc^3 + \sum_{k=1}^m b_k^3 \\
&&&= \sum_{k=1}^m (a_k+c)^3 + \sum_{k=1}^m b_k^3
\end{align*}
\item Note that
\begin{align*}
&& \sum_{k=1}^m a_k^2 + \sum_{k=1}^m (c + b_k)^2 &= \sum_{k=1}^m a_k^2 + mc^2 + 2c \sum_{k=1}^m b_k +\sum_{k=1}^m b_k^2 \\
&&&= \sum_{k=1}^m (a_k+c)^2 + \sum_{k=1}^m b_k^2
\end{align*}
Therefore if we take our balanced subsets of $\{1,2,3,4,5,6,7,8\}$ and take $A \cup (B+8)$ and $B \cup (A+8)$ they will be balanced (including the cubes) so: $A = \{1,4,6,7,10,11,13,16\}, B = \{2,3,5,8,9,12,14,15\}$
\item Notice that:
\begin{align*}
\sum_{a \in A} (n+a)^2 &= |A|n^2 + 2n \sum_{a \in A} a + \sum_{a \in A} a^2 \\
&= 6 n^2 + 2n \sum_{b \in B}b + \sum_{b \in B} b^2 \\
&= n^2 + \sum_{b \in B}(n+b)^2 \\
\end{align*}
Therefore consider the sets $C = \{(n+i)^2 : i \in A\}, D = \{n^2\} \cup \{(n+i)^2 : i \in B\}$
\end{questionparts}
Of the Pure questions, this attracted a relatively small number of responses, although many good solutions were seen. Many candidates were able to write down the partition of the set as their answer to part (i) without much supporting working and this was awarded full marks. Part (ii) was answered well by most candidates, but many responses did not give sufficiently clear explanations. In particular, some simply produced the two binomial expansions and then claimed that the result would be true. A small number of candidates attempted to solve part (iii) without using the result from part (ii). Such attempts were rarely successful. Of those who applied the result from part (ii), many did not show that the properties of a balanced partition would be satisfied by their solution. Some candidates simply showed that these were true by calculating the values for the specific case rather than showing a more general result. In part (iv) many candidates recognised the need to include 0 in the set and then deduced a correct partition. However, in many cases there was insufficient justification that the two sets would have the required property.