There is this summation: $$\sum_{i=10}^{n+10}\sum_{j=i}^{n+10} j$$ and the answer is: $$\frac{1}{3} (n+1) (n+2) (n+15)$$
My question is how do you go from the summation to only the equation with n's.
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please accept my answer if you are satisfied – LinAlg Oct 30 '20 at 00:47
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The inner sum is an arithmetic series. The sum is the average element $0.5(n+10+i)$ times the number of elements ($n+10+1-i$): $$\sum_{j=i}^{n+10} j = 0.5(n+10+i)(n+10+1-i)$$ Plugging this into the summation gives: $$\sum_{i=10}^{n+10}\sum_{j=i}^{n+10} j = \sum_{i=10}^{n+10} 0.5(n+10+1-i)(n+10+i)$$ This is a difficult sum, but you can expand the factors: $$\sum_{i=10}^{n+10}\sum_{j=i}^{n+10} j = \sum_{i=10}^{n+10} (0.5n^2+10.5n+55)+0.5\sum_{i=10}^{n+10}i-0.5\sum_{i=10}^{n+10}i^2$$ The first term is $(n+1)(0.5n^2 + 10.5n +55)$. The second term is again an arithmetic sequence that simplifies to $0.25(n+10+10)(n+10+1-10)$. For the last term you can use this result to get: $$-0.5\left(\frac{1}{6}(n+10)(n+10+1)(2(n+10)+1) - \frac{1}{6}9(9+1)(2\cdot9+1)\right)$$ Now just sum the terms and simplify to obtain your result.
LinAlg
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@BobJoe I first sum from $1$ to $n+10$, and then subtract the first 9 terms – LinAlg Oct 20 '20 at 02:19