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I came across this problem in my textbook: $\sum_{i=10}^{n+10}\sum_{j=i}^{n+10} j$, I know how to do a double summation but not one where they are linked at the starting points like this. The textbook says to write this summation as a function of n. It says to write this summation as a function of n, using summations that have already been proven such as $\sum_{i=1}^{n}i$ = $n*(n+1)/2$

Bob Joe
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    Could you express $\sum_{j=12}^{n+10}j$ in closed form as a function of $n$? – Brian M. Scott Oct 19 '20 at 18:48
  • What tools do you have at your disposal? For example, the Hockey-Stick Identity can be used to handle these kinds of formulas compactly. – robjohn Oct 19 '20 at 18:57
  • I think what Brian and I are asking is that you provide some context to allow us to help you. – robjohn Oct 19 '20 at 19:00
  • It says to write this summation as a function of n, using summations that have already been proven such as$ \sum_{i=1}^{n} i$ is $(n)*(n+1)/2$ – Bob Joe Oct 19 '20 at 19:05
  • $$\sum _{i=10}^{n+10} \left(\sum _{j=i}^{n+10} j\right)=\frac{1}{3} (n+1) (n+2) (n+15)$$ – Raffaele Oct 19 '20 at 20:15
  • @BobJoe: The proper place for context is in the question. Often people who flag or vote-to-close because of lack of context, don't look at the comments. Mentioning what formulas you have proven that might be applicable would also be useful. – robjohn Oct 19 '20 at 20:40
  • @Raffaele what were the steps to get the final answer! Thank you in advance! – Bob Joe Oct 19 '20 at 21:24

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The sum of an arithmetic progression is here $$\sum _{j=\text{start}}^{\text{end}} j=\frac{1}{2} (\text{end}-\text{start}+1) (\text{end}+\text{start})$$ Thus from the inner sum you get $$\sum _{j=i}^{n+10} j=\frac{1}{2} (11-i+n) (10+i+n)=-\frac{i^2}{2}+\frac{i}{2}+\frac{n^2}{2}+\frac{21 n}{2}+55$$ To compute the outer sum we have to compute the following $$\sum _{i=10}^{n+10} \left(\frac{n^2}{2}+\frac{21 n}{2}+55\right)+\frac{1}{2} \sum _{i=10}^{n+10} i-\frac{1}{2} \sum _{i=10}^{n+10} i^2$$ The first sum is simply $(n+1)$ times the argument $$\sum _{i=10}^{n+10} \left(\frac{n^2}{2}+\frac{21 n}{2}+55\right)=(n+1) \left(\frac{n^2}{2}+\frac{21 n}{2}+55\right)$$ The second can be solved with the formula above $$\frac{1}{2} \sum _{i=10}^{n+10} i=\frac{1}{4} (n+1) (n+20)$$ For the third sum we need the formula to sum squares $$\sum _{k=1}^n k^2=\frac{1}{6} n (n+1) (2 n+1)$$ Notice that $$\sum _{i=10}^{n+10} i^2=\sum _{i=1}^{n+10} i^2-\sum _{i=1}^9 i^2$$ Thus

$$\sum _{i=10}^{n+10} i^2=\frac{1}{6} (n+10) (n+11) (2 n+21)-285=\frac{1}{6} (n+1) \left(2 n^2+61 n+600\right)$$ Putting all together

$$\sum _{j=i}^{n+10} j=(n+1) \left(\frac{n^2}{2}+\frac{21 n}{2}+55\right)+\frac{1}{4} (n+1) (n+20)-\frac{1}{12} (n+1) \left(2 n^2+61 n+600\right)=\\=(n+1) \left(\left(\frac{n^2}{2}+\frac{21 n}{2}+55\right)-\frac{1}{12} \left(2 n^2+61 n+600\right)+\frac{n+20}{4}\right)=\\=(n+1)\left(\frac{n^2}{3}+\frac{17 n}{3}+10\right)=\frac{1}{3} (n+1)(n+2) (n+15)$$ And finally we get $$\sum_{i=10}^{n+10}\sum_{j=i}^{n+10} j=\frac{1}{3} (n+1)(n+2) (n+15)$$ This is the reason why using a CAS can be useful and time saving :)

Raffaele
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