STEP I 1994 question 7 solution

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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:

$\displaystyle \sum _{r = (n^{2} + 1)} ^{(n^{2} + 2n + 1)} r \\ = \frac{ (n^{2} + 2n + 1) }{2} \left( n^{2} + 2n + 2 \right) - \frac{ (n^{2} + 1) }{2} \left( n^{2} + 2 \right) + (n^{2} + 1) \\ = \frac{n^{4} + 2n^{3} + 2n^{2} + 2n^{3} + 4n^{2} + 4n + n^{2} + 2n + 2 - n^{4} - 2n^{2} - n^{2} - 2}{2} + (n^{2} + 1) \\ = \frac{4n^{3} + 4n^{2} + 6n}{2} + (n^{2} + 1) \\ = 2n^{3} + 3n^{2} + 3n + 1 \\ = n^{3} + (n + 1)^{3}$

$\displaystyle \sum _{r = 1} ^{n^{2}} r = (0 + 1) + (1 + 8) + (8 + 27) + ... + ((n - 1)^{3} + n^{3}) \\ = 0 + 2(1^{3} + 2^{3} + ... + (n - 1)^{3}) + n^{3} \\ = 2 \sum _{r = 1} ^{n} r^{3} - 2n^{3} + n ^{3} = 2 \sum _{r = 1} ^{n} r^{3} - n^{3}$

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

Hence:

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