Summation proof

[Dr.Peterson]

Just to make sure I'm interpreting this correctly, a summation of a summation is the same as the product of two summations, is that right?

Not in general. What is special about my example, that makes it possible to rewrite it that way? You have to be able to say why you can do what you do.

So initially, you rewrite the second summation with the constant 'j' out front then you equate the summation over 'j' from j=0 to 'n' with the summation over 'k' from k=0 to 'n', which gives the double summation on the right.

Yes, one key is that in the inner (second) summation, j is just a constant that can be factored out. Then I have to change the nested summation into a product of two sums (by again factoring out the constant), and only then can I change the name of the variable. Do you see that? I didn't separate those steps for you.

So let's see, we can also say that,

[imath]\displaystyle\sum_{k = 0}^n \sum_{j = 0}^n j²=\sum_{j = 0}^n j²\sum_{k = 0}^n 1=\sum_{k = 0}^n k²\cdot\ (n+1) = (n+1) \sum_{k = 0}^n k²[/imath]

and

[imath]\displaystyle\sum_{k = 0}^n \sum_{j = 0}^n k²=\sum_{k = 0}^n k²\sum_{j = 0}^n 1=\sum_{k = 0}^n k²\cdot\ (n+1) = (n+1) \sum_{k = 0}^n k²[/imath]

Yes, this looks like you are thinking correctly.

 

[Dr Adam]

Many thanks to you