A-LevelFurther MathematicsSummation of series9231/12 · Oct/Nov 2012 · Q48 Marks

Let f(r) = r(r + 1)(r + 2).

Show that f(r) – f(r – 1) = 3r(r + 1).[1]
Hence show that ∑r(r + 1) = ⅓n(n + 1)(n + 2).[2]
Using the standard result for ∑r, deduce that ∑r² = ⅙n(n + 1)(2n + 1).[2]
Find the sum of the series 1² + 2 × 2² + 3² + 2 × 4² + 5² + 2 × 6² + . . . + 2(n − 1)² + n², where n is odd.[3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2012). The mark or mark type is shown in brackets.

[B1] Verifies given result: r(r + 1)(r + 2) – (r −1)r(r + 1) = r(r + 1)(r + 2 − r + 1) = 3r(r +1) (AG)
[M1] Uses method of differences to sum first series: ∑ (from r=1 to n) r(r + 1) = (1/3)[f(n)-f(0)]
[A1] = (1/3)n(n+1)(n+2) (AG) (Award B1 if 'not hence'.)
[M1] Subtracts ∑r to obtain sum of second series: ∑r² = ∑r(r+1) - ∑r = (1/3)n(n+1)(n+2) - (1/2)n(n+1)
[A1] = (1/6)n(n+1)(2n+4-3) = (1/6)n(n+1)(2n+1) (AG)
[M1] Splits series into two series: (1² + 2² + ... + n²) + (2² + 4² + ... + (n −1)²) = ∑ (from r=1 to n) r² + 4 ∑ (from r=1 to (n-1)/2) r²
[M1A1] Applies sum of squares formula to obtain result: (1/6)n(n+1)(2n+1) + (1/6) * 4 * ((n-1)/2) * ((n-1)/2+1) * (2(n-1)/2+1) = (1/2)n²(n+1)

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