Thanks for your quick reply! I'm not sure how to proceed from here..Start with:
(j + k)2 = j2 + k2 + 2*j*k ......................................................................appended later
Please show us what you have tried and exactly where you are stuck.
Please follow the rules of posting in this forum, as enunciated at:
Please share your work/thoughts about this problem
Did you check the RHS and simplify ?
You have a couple of mistakes that I easily caught. That last summation is not k^2*n. Why are you not summing up that middle and last expression over k? In the 1st summation on the 2nd line you are summing up a constant as there are no k's in it. Don't make the same mistake as you did in the other summation of a constant.
Your answer should be in terms of n.Yes I understand that, I could go ahead and sum over remaining k terms but this is just going to get me a longer expression in terms of 'n' which seems to get further away from what I'm aiming for.
I wanted to stop here simply to check if I've already gone wrong somewhere. Something hasn't gone wrong, it's just incomplete
I don't think that is mandatory. Only the equality in op needs to be proven.Your answer should be in terms of n.
Why is everyone saying that \(\displaystyle \displaystyle \sum_{j = 0}^n 1 = n? \) j does not take on exactly n different values![imath]\displaystyle \sum_{k = 0}^n \sum_{j = 0}^n (j + k)^2[/imath]
[imath]\displaystyle = \sum_{k = 0}^n \sum_{j = 0}^n j^2 + 2 \sum_{j = 0}^n \sum_{k = 0}^n jk + \sum_{j = 0}^n \sum_{k = 0}^n k^2[/imath]
[imath]\displaystyle = \sum_{k = 0}^n \dfrac{1}{6} n (n+1) (2n +1) + 2 \sum_{k = 0}^n k \dfrac{1}{2}n (n+1) + \sum_{k = 0}^n k^2 \left ( \sum_{j = 0}^n 1 \right )[/imath]
[imath]\displaystyle = \sum_{k = 0}^n \dfrac{1}{6} n (n+1) (2n +1) + 2 \sum_{k = 0}^n k \dfrac{1}{2}n (n+1) + \sum_{k = 0}^n k^2 n[/imath]
To get to the next step, what do those n's inside the summations have to do with the summations? Nothing! They are just a constant. So pull them out of the sums... they are just a multiplicative factor now.
-Dan
Crap. I typed [imath]\displaystyle \sum_{j = 0}^n[/imath] but I was thinking of summing from j = 1. Good catch!Why is everyone saying that \(\displaystyle \displaystyle \sum_{j = 0}^n 1 = n? \) j does not take on exactly n different values!
[imath]\displaystyle \sum_{k = 0}^n \sum_{j = 0}^n (j + k)^2[/imath]
[imath]\displaystyle = \sum_{k = 0}^n \sum_{j = 0}^n j^2 + 2 \sum_{j = 0}^n \sum_{k = 0}^n jk + \sum_{j = 0}^n \sum_{k = 0}^n k^2[/imath]
[imath]\displaystyle = \sum_{k = 0}^n \dfrac{1}{6} n (n+1) (2n +1) + 2 \sum_{k = 0}^n k \dfrac{1}{2}n (n+1) + \sum_{k = 0}^n k^2 \left ( \sum_{j = 0}^n 1 \right )[/imath]
[imath]\displaystyle = \sum_{k = 0}^n \dfrac{1}{6} n (n+1) (2n +1) + 2 \sum_{k = 0}^n k \dfrac{1}{2}n (n+1) + \sum_{k = 0}^n k^2 n[/imath]
To get to the next step, what do those n's inside the summations have to do with the summations? Nothing! They are just a constant. So pull them out of the sums... they are just a multiplicative factor now.
-Dan
Thanks Dan! However this isn't the same expression as the right hand side of the original 'Show that..' problem, shown here ?[imath]\displaystyle \sum_{k = 0}^n \sum_{j = 0}^n (j + k)^2[/imath]
[imath]\displaystyle = \sum_{k = 0}^n \sum_{j = 0}^n j^2 + 2 \sum_{j = 0}^n \sum_{k = 0}^n jk + \sum_{j = 0}^n \sum_{k = 0}^n k^2[/imath]
[imath]\displaystyle = \sum_{k = 0}^n \dfrac{1}{6} n (n+1) (2n +1) + 2 \sum_{k = 0}^n k \dfrac{1}{2}n (n+1) + \sum_{k = 0}^n k^2 \left ( \sum_{j = 0}^n 1 \right )[/imath]
[imath]\displaystyle = \sum_{k = 0}^n \dfrac{1}{6} n (n+1) (2n +1) + 2 \sum_{k = 0}^n k \dfrac{1}{2}n (n+1) + \sum_{k = 0}^n k^2 n[/imath]
To get to the next step, what do those n's inside the summations have to do with the summations? Nothing! They are just a constant. So pull them out of the sums... they are just a multiplicative factor now.
-Dan
Now work with the RHS of the equation.Thanks Dan! However this isn't the same expression as the right hand side of the original 'Show that..' problem, shown here ?
View attachment 34223
I didn't finish the problem. I wrote my post to show you places where you went wrong in the work that you showed. As Subhotosh Khan mentioned, you now need to work on the RHS of your expression to see if they match.Thanks Dan! However this isn't the same expression as the right hand side of the original 'Show that..' problem, shown here ?
View attachment 34223
You really don't have to go as far as topsquark went, if you pay attention to the RHS as you do it. After this part,Thanks Dan! However this isn't the same expression as the right hand side of the original 'Show that..' problem, shown here ?
View attachment 34223
Because my suggestion is far easier. I almost finished it; it will just take another line or two of similar work.As I stated before, why not do this by induction?
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?You really don't have to go as far as topsquark went, if you pay attention to the RHS as you do it. After this part,
[imath]\displaystyle \sum_{k = 0}^n \sum_{j = 0}^n (j + k)^2 = \sum_{k = 0}^n \sum_{j = 0}^n j^2 + 2 \sum_{j = 0}^n \sum_{k = 0}^n jk + \sum_{j = 0}^n \sum_{k = 0}^n k^2[/imath]
you can rewrite some of the summations to produce what the RHS has.
For example, [imath]\displaystyle\sum_{j = 0}^n \sum_{k = 0}^n jk=\sum_{j = 0}^n j\sum_{k = 0}^n k=\sum_{k = 0}^n k\cdot\sum_{k = 0}^n k[/imath]
Do you see why?