The standard result is sum_r=1^m r^3 = [m(m+1)/2]^2. Which expression gives sum_r=k+1^3k r^3?

The standard result is $\sum_{r=1}^{m} r^3 = \left[\frac{m(m+1)}{2}\right]^2$ . Which expression gives $\sum_{r=k+1}^{3k} r^3$ ?

\left[\frac{3k(3k+1)}{2}\right]^2-\left[\frac{(k+1)(k+2)}{2}\right]^2

\left[\frac{3k(3k+1)}{2}-\frac{k(k+1)}{2}\right]^2

\left[\frac{3k(3k+1)}{2}\right]^2-\left[\frac{k(k+1)}{2}\right]^2

\left[\frac{3k(3k+1)}{2}\right]^2+\left[\frac{k(k+1)}{2}\right]^2

Explanation

A sum running from

r=k+1

is the sum as far as the top limit less the sum as far as

k

, not as far as

k+1

. So

\sum_{r=k+1}^{3k}r^3=\left[\frac{3k(3k+1)}{2}\right]^2-\left[\frac{k(k+1)}{2}\right]^2=\frac{k^2}{4}\left[9(3k+1)^2-(k+1)^2\right]=\frac{k^2}{4}\left(80k^2+52k+8\right)=k^2(4k+1)(5k+2)

Derived from ZIMSEC Further Mathematics 9187/1, November 2010, Q3

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