Double Summation: sum[i=1, m] sum[j=1, n] (i + j^2)

[xwy]

Hi guys. Would appreciate if anyone of you could help me with this question.

\(\displaystyle \mbox{Question: Compute and expand: }\, \)\(\displaystyle \displaystyle \sum_{i=1}^m\, \sum_{j=1}^n\, (i\, +\, j^2)\)

I know that these 2 rules:

. . . . .\(\displaystyle \mbox{1) }\, \)\(\displaystyle \displaystyle \sum_{i=1}^n\, i^2\, =\, 1^2\, +\, 2^2\, +\, 3^2\, +\, ...\, +\, n^2\, =\, \dfrac{1}{6}\, n\,(n\, +\, 1)\,(2n\, +\, 1)\)

. . . . .\(\displaystyle \mbox{2) }\, \)\(\displaystyle \displaystyle \sum_{i=1}^n\, i\, =\, 1\, +\, 2\, +\, 3\, +\, ... \, +\, n\, =\, \dfrac{1}{2}\,n\,(n\, +\, 1)\)

Thus, substituting in:

. . . . .\(\displaystyle \displaystyle \sum_{i=1}^m \, i\, +\, \dfrac{1}{6}\, n\,(n\, +\, 1)\,(2n\, +\, 1)\)

. . . . .\(\displaystyle \displaystyle \sum_{i=1}^m \, i\, +\, \dfrac{1}{3}\, n^3\, +\, \dfrac{1}{2}\, n^2\, +\, \dfrac{1}{2}\, n\)

. . . . .\(\displaystyle \dfrac{1}{2}\, m\, (m\, +\, 1)\, +\, \dfrac{1}{3}\, n^3\, +\, \dfrac{1}{2}\, n^2\, +\, \dfrac{1}{2}\, n\)

. . . . .\(\displaystyle \dfrac{1}{2}\, m^2\, +\, \dfrac{1}{2}\, m\, +\, \dfrac{1}{3}\, n^3\, +\, \dfrac{1}{2}\, n^2\, +\, \dfrac{1}{2}\, n\)

And I am stuck here.

The answer should be:

. . . . .\(\displaystyle \dfrac{1}{6}\, mn\, (2n^2\, +\,3n\, +\, 3m\, +\, 4)\)

Thanks. I have a hard time doing it!

 

[pka]

Hi guys. Would appreciate if anyone of you could help me with this question.

\(\displaystyle \mbox{Question: Compute and expand: }\, \)\(\displaystyle \displaystyle \sum_{i=1}^m\, \sum_{j=1}^n\, (i\, +\, j^2)\)

I know that these 2 rules:

. . . . .\(\displaystyle \mbox{1) }\, \)\(\displaystyle \displaystyle \sum_{i=1}^n\, i^2\, =\, 1^2\, +\, 2^2\, +\, 3^2\, +\, ...\, +\, n^2\, =\, \dfrac{1}{6}\, n\,(n\, +\, 1)\,(2n\, +\, 1)\)

. . . . .\(\displaystyle \mbox{2) }\, \)\(\displaystyle \displaystyle \sum_{i=1}^n\, i\, =\, 1\, +\, 2\, +\, 3\, +\, ... \, +\, n\, =\, \dfrac{1}{2}\,n\,(n\, +\, 1)\)

Thus, substituting in:

. . . . .\(\displaystyle \displaystyle \sum_{i=1}^m \, i\, +\, \dfrac{1}{6}\, n\,(n\, +\, 1)\,(2n\, +\, 1)\)

. . . . .\(\displaystyle \displaystyle \sum_{i=1}^m \, i\, +\, \dfrac{1}{3}\, n^3\, +\, \dfrac{1}{2}\, n^2\, +\, \dfrac{1}{2}\, n\)

. . . . .\(\displaystyle \dfrac{1}{2}\, m\, (m\, +\, 1)\, +\, \dfrac{1}{3}\, n^3\, +\, \dfrac{1}{2}\, n^2\, +\, \dfrac{1}{2}\, n\)

. . . . .\(\displaystyle \dfrac{1}{2}\, m^2\, +\, \dfrac{1}{2}\, m\, +\, \dfrac{1}{3}\, n^3\, +\, \dfrac{1}{2}\, n^2\, +\, \dfrac{1}{2}\, n\)

And I am stuck here.

The answer should be:

. . . . .\(\displaystyle \dfrac{1}{6}\, mn\, (2n^2\, +\,3n\, +\, 3m\, +\, 4)\)

Thanks. I have a hard time doing it!

\(\displaystyle \begin{align*} \sum\limits_{k = 1}^m {\left[ {\sum\limits_{j = 1}^n {\left( {k + {j^2}} \right)} } \right]} &= \sum\limits_{k = 1}^m {\left[ {nk + \frac{{n(n + 1)(2n + 1)}}{6}} \right]} \\ &= n\frac{{m(m + 1)}}{2} + m\frac{{n(n + 1)(2n + 1)}}{6}\\ &= \frac{1}{6}\left[ {3nm + nm(n + 1)(2n + 1)} \right] \end{align*}\)
Can you finish? See here.

 

[Ishuda]

Hi guys. Would appreciate if anyone of you could help me with this question.

\(\displaystyle \mbox{Question: Compute and expand: }\, \)\(\displaystyle \displaystyle \sum_{i=1}^m\, \sum_{j=1}^n\, (i\, +\, j^2)\)

I know that these 2 rules:

. . . . .\(\displaystyle \mbox{1) }\, \)\(\displaystyle \displaystyle \sum_{i=1}^n\, i^2\, =\, 1^2\, +\, 2^2\, +\, 3^2\, +\, ...\, +\, n^2\, =\, \dfrac{1}{6}\, n\,(n\, +\, 1)\,(2n\, +\, 1)\)

. . . . .\(\displaystyle \mbox{2) }\, \)\(\displaystyle \displaystyle \sum_{i=1}^n\, i\, =\, 1\, +\, 2\, +\, 3\, +\, ... \, +\, n\, =\, \dfrac{1}{2}\,n\,(n\, +\, 1)\)

Thus, substituting in:

. . . . .\(\displaystyle \displaystyle \sum_{i=1}^m \, i\, +\, \dfrac{1}{6}\, n\,(n\, +\, 1)\,(2n\, +\, 1)\)

. . . . .\(\displaystyle \displaystyle \sum_{i=1}^m \, i\, +\, \dfrac{1}{3}\, n^3\, +\, \dfrac{1}{2}\, n^2\, +\, \dfrac{1}{2}\, n\)

. . . . .\(\displaystyle \dfrac{1}{2}\, m\, (m\, +\, 1)\, +\, \dfrac{1}{3}\, n^3\, +\, \dfrac{1}{2}\, n^2\, +\, \dfrac{1}{2}\, n\)

. . . . .\(\displaystyle \dfrac{1}{2}\, m^2\, +\, \dfrac{1}{2}\, m\, +\, \dfrac{1}{3}\, n^3\, +\, \dfrac{1}{2}\, n^2\, +\, \dfrac{1}{2}\, n\)

And I am stuck here.

The answer should be:

. . . . .\(\displaystyle \dfrac{1}{6}\, mn\, (2n^2\, +\,3n\, +\, 3m\, +\, 4)\)

Thanks. I have a hard time doing it!

As pka has implied, maybe a couple of other rules you should learn
(3) \(\displaystyle \sum\limits_{j = 1}^n\, k\, a_j\, =\, k\, \sum\limits_{j = 1}^n\, a_j\)
(4) \(\displaystyle \sum\limits_{j = 1}^n\, [ a_j\, +\, b_j ] = \sum\limits_{j = 1}^n\, a_j\, +\, \sum\limits_{j = 1}^n\, b_j\)

That is, (3) if you have something inside the summation which is independent of the summation index, it can be moved outside the summation and (4) summation is distributive over addition.