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STEP I 1994 question 7 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1994 question 7 solution

To prove: (n^{2} + 1) + (n^{2} + 2) + ... + (n + 1)^{2} = n^{3} + (n + 1)^{3}

The summation is of the integers from (n^{2} + 1) to (n + 1)^{2} , hence:

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\displaystyle \sum _{r = 1} ^{n^{2}} r = \frac{n^{2}}{2} (n^{2} + 1) = \frac{n^{4} + n^{2}}{2}

Hence:

\displaystyle 2 \sum _{r = 1} ^{n} r^{3} - n^{3} = \frac{n^{4} + n^{2}}{2} \\ \implies \sum _{r = 1} ^{n} r^{3} = \frac{n^{4} + 2n^{3} + n^{2}}{4} \\ = \frac{n^{2}(n^{2} + 2n + 1)}{4} \\ = \frac{1}{4} n^{2} (n + 1)^{2} .

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