How to Solve a Double Summation Problem?

[mexion]

Homework Statement
\sum_{k=1}^{3} \sum_{j=0}^{4}k^{j}

The attempt at a solution
so this above means:
(1+1^{1}+1^{2}+1^{3}+1^{4})*(1+2^{1}+2^{2}...)*(1+3...)*...
or
(1*1^{1}*1^{2}*1^{3}*1^{4})+(1*2^{1}*2^{2}...)+(1*3...)+...

 

[kobjob]

It doesn't make a difference which order you do it in. You can see if you type the following into matlab:

syms k
syms j

symsum(symsum(k^j,j,0,4),k,1,3)

or,

symsum(symsum(k^j,k,1,3),j,0,4)

the answer is 157 either way.

 

[mexion]

so 1 more question
\sum_{k=0}^{4} \sum_{j=1}^{5} (3^{k} + jk)
how to calculate this without Matlab.
Should I transform it? (but have no idea how)
or just have to put [0;4] for k, and [1;5] for j, and add everything ?

 

[kobjob]

You don't need to do a transform, if you start with the sum over j you will get

\sum_{k=0}^{4} (3^k+1k+ ... 3^k+5k) = 0+ ...+ 3^4+20

 

[Ray Vickson]

Homework Statement
\sum_{k=1}^{3} \sum_{j=0}^{4}k^{j}

The attempt at a solution
so this above means:
(1+1^{1}+1^{2}+1^{3}+1^{4})*(1+2^{1}+2^{2}...)*(1+3...)*...
or
(1*1^{1}*1^{2}*1^{3}*1^{4})+(1*2^{1}*2^{2}...)+(1*3...)+...

It means the second one (which = 157); the first would be product_{k=1..3} sum_{j=0..4} k^j = 18755.

RGV