Are these two sums equal?$\sum\limits_{i=1}^{n^2}\sum\limits_{j=i}^{i+n}\sum\limits_{k=1}^j = \sum\limits_{i=1}^{n^2}i+(i+1)+(i+1)+...+(i+n)$

Question

$\sum\limits_{i=1}^{n^2}\sum\limits_{j=i}^{i+n}\sum\limits_{k=1}^j = \sum\limits_{i=1}^{n^2}i+(i+1)+(i+1)+...+(i+n)$

Are these two sums equal? If n = 2 so the result of $\sum\limits_{i=1}^{n^2}\sum\limits_{j=i}^{i+n}\sum\limits_{k=1}^j$ is equal 30 and for $\sum\limits_{i=1}^{n^2}i+(i+1)+(i+1)+...+(i+n)$ is equal 42 !

But in the book wrote two formula is equal! Please, description why if I have wrong!

I know the simpled formula is as: $$\sum\limits_{j=i}^{i+n}\sum\limits_{k=1}^j=\sum\limits_{j=i}^{i+n}j= i+(i+1)+(i+1)+…+(i+n)$$

For the result 30 see :

user · Accepted Answer · 2019-12-07T12:20:04.843

1

We have that

$$\sum\limits_{k=1}^j1=j$$

and then

$$\sum\limits_{j=i}^{i+n}\sum\limits_{k=1}^j=\sum\limits_{j=i}^{i+n}j= i+(i+1)+(i+1)+…+(i+n)$$

therefore the equality holds, you have probably made some mistake in the calculation.

We have that for $n=2$

$$\sum\limits_{i=1}^{4}\sum\limits_{j=i}^{i+2}\sum\limits_{k=1}^j = \sum\limits_{i=1}^{4}\sum\limits_{j=i}^{i+2}j= \sum\limits_{i=1}^{4} (i+i+1+i+2)=\sum\limits_{i=1}^{4}( 3i+3)=$$$$=3\sum\limits_{i=1}^{4}i+3\sum\limits_{i=1}^{4}1=30+12=42$$

edited Dec 07 '19 at 12:20

answered Dec 07 '19 at 12:10

user

154,566

Yes ... I know this ... but please give n =2 and get the result ! please edit it and write for n=2. tanks! – Michael Dec 07 '19 at 12:13
@Michael I obtain indeed $42$. How did you obtain $30$? – user Dec 07 '19 at 12:22
Please reload the page! the question is edited. – Michael Dec 07 '19 at 12:34
@Michael We are summing $\sum\limits_{j=i}^{i+2}j$ that is for $$i=1\implies \sum\limits_{j=1}^{3}j=6$$ $$i=2\implies \sum\limits_{j=2}^{4}j=9$$ $$i=3\implies \sum\limits_{j=3}^{5}j=12$$ $$i=4\implies \sum\limits_{j=4}^{6}j=15$$ – user Dec 07 '19 at 12:39
thanks .. please write i=2⟹∑j=24j=9 (i=2) for example to for me. – Michael Dec 07 '19 at 12:46
@Michael Sorry I didn't understand what you are asking for. – user Dec 07 '19 at 12:48
Sorry, I bad wrote! write as complete how to get 9 to i =2 . – Michael Dec 07 '19 at 12:49
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@Michael $$\sum\limits_{j=2}^{4}j=2+3+4=9$$ – user Dec 07 '19 at 12:51
tanks!!!!!!!!!!!!! – Michael Dec 07 '19 at 12:52

Are these two sums equal?$\sum\limits_{i=1}^{n^2}\sum\limits_{j=i}^{i+n}\sum\limits_{k=1}^j = \sum\limits_{i=1}^{n^2}i+(i+1)+(i+1)+...+(i+n)$

1 Answers1