Moderate -0.5 This is a straightforward application of standard summation formulae with basic algebraic manipulation. Students need only split the sum, apply three given formulae, and simplify—no problem-solving insight required. While it's Further Maths content, it's a routine textbook exercise that's actually easier than many single-maths proof questions.
5 Use the standard results for \(\sum _ { r = 1 } ^ { n } r , \sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\) to show that, for all positive integers \(n\),
$$\sum _ { r = 1 } ^ { n } \left( 8 r ^ { 3 } - 6 r ^ { 2 } + 2 r \right) = 2 n ^ { 3 } ( n + 1 )$$
Attempt to factorise or expand and simplify; Obtain given answer correctly
Total: 6 marks
$8\sum r^3, -6\sum r^2, 2\sum r$ | M1 | Consider the sum of three separate terms
$8\sum r^3 = 2n^2(n+1)^2$ | A1 | Correct formula stated or used a.e.f.
$6\sum r^2 = n(n+1)(2n+1)$ | A1 | Correct formula stated or used a.e.f.
$2\sum r = n(n+1)$ | A1 | Correct term seen
$2n^3(n+1)$ | M1 A1 | Attempt to factorise or expand and simplify; Obtain given answer correctly
**Total: 6 marks**
5 Use the standard results for $\sum _ { r = 1 } ^ { n } r , \sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r ^ { 3 }$ to show that, for all positive integers $n$,
$$\sum _ { r = 1 } ^ { n } \left( 8 r ^ { 3 } - 6 r ^ { 2 } + 2 r \right) = 2 n ^ { 3 } ( n + 1 )$$
\hfill \mbox{\textit{OCR FP1 2006 Q5 [6]}}