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I am unsure of how to proceed about finding the solution to this problem. $$\sum_{i=6}^8(\sum_{j=i}^8 (j+1)^2)$$

Obviously the last step is not to difficult, but the fact that the lower limit for the summation in brackets is i I am not sure how to solve this. In classes so far we have only really dealt with cases where j=1. I assume however it would be beneficial to reindex, so that we have

$$\sum_{i=6}^8(\sum_{k=i+1}^9 k^2)$$

Any help is greatly appreciated!

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Since there are only three values of $i$, I'd rewrite this as $$\sum\limits_{j=6}^8 (j+1)^2 + \sum\limits_{j=7}^8 (j+1)^2 + \sum\limits_{j=8}^8 (j+1)^2$$

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As pointed out in the first answer, it would be easier to compute it by simple expansion since $i$ can take on only $3$ values.

However since you asked about reindexing this is a possible approach:

$$\sum_{i=6}^8\sum_{j=i}^8(j+1)^2=\sum_{i=7}^9\sum_{j=i}^9 j^2=\sum_{7\le i\le j\le 9} j^2=\sum_{j=7}^9\sum_{i=7}^j j^2=\sum_{j=7}^9(j-6)j^2=7^2+2(8^2)+3(9^2)\;\;\blacksquare$$

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