By considering \(( 2 r + 1 ) ^ { 2 } - ( 2 r - 1 ) ^ { 2 }\), use the method of differences to prove that
$$\sum _ { r = 1 } ^ { n } r = \frac { 1 } { 2 } n ( n + 1 )$$
By considering \(( 2 r + 1 ) ^ { 4 } - ( 2 r - 1 ) ^ { 4 }\), use the method of differences and the result given in part (i) to prove that
$$\sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }$$
The sums \(S\) and \(T\) are defined as follows:
$$\begin{aligned}
& S = 1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } + 4 ^ { 3 } + \ldots + ( 2 N ) ^ { 3 } + ( 2 N + 1 ) ^ { 3 } , \\
& T = 1 ^ { 3 } + 3 ^ { 3 } + 5 ^ { 3 } + 7 ^ { 3 } + \ldots + ( 2 N - 1 ) ^ { 3 } + ( 2 N + 1 ) ^ { 3 } .
\end{aligned}$$
Use the result given in part (ii) to show that \(S = ( 2 N + 1 ) ^ { 2 } ( N + 1 ) ^ { 2 }\).
Hence, or otherwise, find an expression in terms of \(N\) for \(T\), factorising your answer as far as possible.
Deduce the value of \(\frac { S } { T }\) as \(N \rightarrow \infty\).
(i) By considering $( 2 r + 1 ) ^ { 2 } - ( 2 r - 1 ) ^ { 2 }$, use the method of differences to prove that
$$\sum _ { r = 1 } ^ { n } r = \frac { 1 } { 2 } n ( n + 1 )$$
(ii) By considering $( 2 r + 1 ) ^ { 4 } - ( 2 r - 1 ) ^ { 4 }$, use the method of differences and the result given in part (i) to prove that
$$\sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }$$
The sums $S$ and $T$ are defined as follows:
$$\begin{aligned}
& S = 1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } + 4 ^ { 3 } + \ldots + ( 2 N ) ^ { 3 } + ( 2 N + 1 ) ^ { 3 } , \\
& T = 1 ^ { 3 } + 3 ^ { 3 } + 5 ^ { 3 } + 7 ^ { 3 } + \ldots + ( 2 N - 1 ) ^ { 3 } + ( 2 N + 1 ) ^ { 3 } .
\end{aligned}$$
(iii) Use the result given in part (ii) to show that $S = ( 2 N + 1 ) ^ { 2 } ( N + 1 ) ^ { 2 }$.\\
(iv) Hence, or otherwise, find an expression in terms of $N$ for $T$, factorising your answer as far as possible.\\
(v) Deduce the value of $\frac { S } { T }$ as $N \rightarrow \infty$.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q11 EITHER}}