Summation proof

Dr Adam said:
Show that,
View attachment 34201

Thanks in advance!
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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
 
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By mathematical induction.

1st: let n=0: Are both sides equal? They should be
2nd: let n=k: Assume both sides are equal.
3rd: let n=k+1: Using the assumption, if necessary, show that both sides are equal.
 
Subhotosh Khan said:
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
Click to expand...
Thanks for your quick reply! I'm not sure how to proceed from here..

1664588558824.png
 
Subhotosh Khan said:
Did you check the RHS and simplify ?
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Yes but when I simplify I end up with long expression in terms of 'n' when I'm trying to get it in terms of summation over 'k'.

So I've got this far but I don't know how to express this 'n heavy' expression in terms of 'k'.

Can you help?

1664592082874.png
 
Dr Adam said:
View attachment 34204
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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.

j and k is are dummy variables. This means your final answer should not have j's and k's in it. In your last line after you (claim to have) finished up some of the summations you still have k's in it. That should tell you that something went wrong!
 
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
 
[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
 
Dr Adam said:
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
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Your answer should be in terms of n.
 
topsquark said:
[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
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Why is everyone saying that \(\displaystyle \displaystyle \sum_{j = 0}^n 1 = n? \) j does not take on exactly n different values!
 
Steven G said:
Why is everyone saying that \(\displaystyle \displaystyle \sum_{j = 0}^n 1 = n? \) j does not take on exactly n different values!
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Crap. I typed [imath]\displaystyle \sum_{j = 0}^n[/imath] but I was thinking of summing from j = 1. Good catch!

-Dan
 
topsquark said:
[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
Click to expand...
topsquark said:
[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
Click to expand...
Thanks Dan! However this isn't the same expression as the right hand side of the original 'Show that..' problem, shown here ?

1664755321538.png
 
Dr Adam said:
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
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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.

-Dan
 
Dr Adam said:
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
Click to expand...
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?
 
Steven G said:
As I stated before, why not do this by induction?
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Because my suggestion is far easier. I almost finished it; it will just take another line or two of similar work.

But certainly induction will work ... eventually.
 
Dr.Peterson said:
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?
Click to expand...
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?

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.

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]
 
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