Moderate -0.5 This is a straightforward application of standard summation formulae requiring expansion of r(r+1) into r² + r, substitution of known formulae, and algebraic simplification. While it's a Further Maths topic, it's a routine textbook exercise with clear guidance ('use formulae') and no problem-solving insight required, making it slightly easier than an average A-level question.
1 Use formulae for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) to show that
$$\sum _ { r = 1 } ^ { n } r ( r + 1 ) = \frac { 1 } { 3 } n ( n + 1 ) ( n + 2 )$$
For either correct sum formula stated / For completely correct expression / For factoring attempt / For showing given answer correctly
Total: 5 marks
$\sum_{r=1}^{n} r(r+1) = \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r = \frac{1}{6}n(n+1)(2n+1) + \frac{1}{2}n(n+1)$ | M1 | For considering the two separate sums
$= \frac{1}{6}n(n+1)(2n+3) = \frac{1}{3}n(n+1)(n+2)$ | A1, A1, M1, A1 | For either correct sum formula stated / For completely correct expression / For factoring attempt / For showing given answer correctly
**Total: 5 marks**
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1 Use formulae for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ to show that
$$\sum _ { r = 1 } ^ { n } r ( r + 1 ) = \frac { 1 } { 3 } n ( n + 1 ) ( n + 2 )$$
\hfill \mbox{\textit{OCR FP1 Q1 [5]}}