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Algebra Difficulty 5.1 AIME, harder Find the answer

Find the value of ∑k=160∑n=1kn261−2n\sum_{k=1}^{60} \sum_{n=1}^{k} \frac{n^{2}}{61-2 n}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Change the order of summation and simplify the inner sum: ∑k=160∑n=1kn261−2n=∑n=160∑k=n60n261−2n=∑n=160n2(61−n)61−2n\sum_{k=1}^{60} \sum_{n=1}^{k} \frac{n^{2}}{61-2 n} =\sum_{n=1}^{60} \sum_{k=n}^{60} \frac{n^{2}}{61-2 n} =\sum_{n=1}^{60} \frac{n^{2}(61-n)}{61-2 n}. Then, we rearrange the sum to add the terms corresponding to nn and 61−n61-n: ∑n=160n2(61−n)61−2n=∑n=130(n2(61−n)61−2n+(61−n)2(61−(61−n))61−2(61−n))=∑n=130n2(61−n)−n(61−n)261−2n=∑n=130n(61−n)(n−(61−n))61−2n=∑n=130−n(61−n)=∑n=130n2−61n\sum_{n=1}^{60} \frac{n^{2}(61-n)}{61-2 n} =\sum_{n=1}^{30}\left(\frac{n^{2}(61-n)}{61-2 n}+\frac{(61-n)^{2}(61-(61-n))}{61-2(61-n)}\right) =\sum_{n=1}^{30} \frac{n^{2}(61-n)-n(61-n)^{2}}{61-2 n} =\sum_{n=1}^{30} \frac{n(61-n)(n-(61-n))}{61-2 n} =\sum_{n=1}^{30}-n(61-n) =\sum_{n=1}^{30} n^{2}-61 n. Finally, using the formulas for the sum of the first kk squares and sum of the first kk positive integers, we conclude that this last sum is 30(31)(61)6−6130(31)2=−18910\frac{30(31)(61)}{6}-61 \frac{30(31)}{2}=-18910. So, the original sum evaluates to -18910.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.