Moderate -0.3 This is a guided, multi-part question that walks students through deriving the sum of squares formula using the method of differences. Part (i) is straightforward telescoping, part (ii) is simple algebraic expansion, and part (iii) combines these results with a given formula. While it requires understanding the method of differences, the question provides clear scaffolding and involves only routine algebraic manipulation at each step, making it slightly easier than average.
5. (i) Use the method of differences to show that
$$\sum _ { r = 1 } ^ { n } \left\{ ( r + 1 ) ^ { 3 } - r ^ { 3 } \right\} = ( n + 1 ) ^ { 3 } - 1$$
(ii) Show that \(( r + 1 ) ^ { 3 } - r ^ { 3 } \equiv 3 r ^ { 2 } + 3 r + 1\).
(iii) Use the results in parts (i) and (ii) and the standard result for \(\sum _ { r = 1 } ^ { n } r\) to show that
$$3 \sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 2 } n ( n + 1 ) ( 2 n + 1 )$$
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5. (i) Use the method of differences to show that
$$\sum _ { r = 1 } ^ { n } \left\{ ( r + 1 ) ^ { 3 } - r ^ { 3 } \right\} = ( n + 1 ) ^ { 3 } - 1$$
(ii) Show that $( r + 1 ) ^ { 3 } - r ^ { 3 } \equiv 3 r ^ { 2 } + 3 r + 1$.\\
(iii) Use the results in parts (i) and (ii) and the standard result for $\sum _ { r = 1 } ^ { n } r$ to show that
$$3 \sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 2 } n ( n + 1 ) ( 2 n + 1 )$$
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\hfill \mbox{\textit{SPS SPS FM Pure 2023 Q5 [10]}}