Surprising Summation (3): $\sum_{i=1}^n\sum_{j=1}^i 2(n-i)+1=\sum_{i=1}^n i^2$

Question

Show that

$$\sum_{i=1}^n\sum_{j=1}^i 2(n-i)+1=\sum_{i=1}^ni^2$$

without expanding the summation to its closed-form solution, i.e. $\dfrac 16n(n+1)(2n+1)$ or equivalent.

Background as requested:
The summands for both equations are not the same but the results are the same. The challenge here is to transform LHS into RHS without solving the summation. It seems like an interesting challenge. If you like this question, please vote to reopen it!

In one of your summations, you are summing over $j$ but have no $j$ index term.. — Matthew Cassell, Aug 24 '15 at 17:16
You're right, it doesn't. I shouldn't try and do maths when tired. — Matthew Cassell, Aug 24 '15 at 17:22

score 2 · Accepted Answer · edited Aug 25 '15 at 12:02

2

\begin{array}\\ \sum_{i=1}^n\sum_{j=1}^i (2(n-i)+1) &=\sum_{j=1}^n\sum_{i=j}^n (2(n-i)+1) \qquad \text{(switch order of summation)}\\ &=\sum_{j=1}^n\sum_{i=0}^{n-j} (2i+1) \qquad (i \to n-i)\\ &=\sum_{j=1}^n(n-j+1)^2 \qquad\text{(since }\sum_{j=0}^m (2j+1) = (m+1)^2)\\ &=\sum_{j=1}^n j^2 \qquad (j \to n-j+1)\\ \end{array}

edited Aug 25 '15 at 12:02

Vim

13,640

answered Aug 24 '15 at 18:30

marty cohen

107,799

That's exactly the way I would have proceeded!! +1 – Mark Viola Aug 24 '15 at 22:40
Thanks for your solution. Very nice approach (+1). I used the substitution of $(n-i+1)$ but it's similar. The last step requires knowing that the sum of the first $m$ odd numbers equals $m^2$. Is it possible to modify the answer so that it does not require this prior knowledge? The objective is to find an "empirical" proof using only first principles. – Hypergeometricx Aug 25 '15 at 08:02
Accepting this answer as it works its way from LHS to RHS without first knowing what RHS is, and without expanding to the known result of any summation. – Hypergeometricx Aug 25 '15 at 15:38

score 1 · Answer 2 · answered Aug 24 '15 at 17:47

1

$$\sum_{i=1}^{n}\sum_{j=1}^{i}(2(n-i)+1)-\sum_{i=1}^{n}i^2$$ $$=\sum_{i=1}^{n}(2n+1)i-3\sum_{i=1}^{n}i^2$$ $$=(2n+1)\frac n2(n+1)-3\times\frac n6(n+1)(2n+1)=0$$

answered Aug 24 '15 at 17:47

David Quinn

34,121

Thanks for your solution. The question did specify not using expansion though. Would you be able to modify your answer taking this into consideration? – Hypergeometricx Aug 25 '15 at 07:59
what do you mean by expansion exactly, and where has this occurred? – David Quinn Aug 25 '15 at 08:45
Expansion of the summation meaning using the formulas for $\sum_{i=1}^ni$ and $\sum_{i=1}^n i^2$. – Hypergeometricx Aug 25 '15 at 08:46
so you want to arrive at the result by manipulation of summation limits only? – David Quinn Aug 25 '15 at 08:50
Yes, or other similar methods, as long as no expansion is involved. Or perhaps the term should be closed-form solution. – Hypergeometricx Aug 25 '15 at 08:52
not sure what you mean by the last sentence. have you thought of maybe rephrasing your question to make it more specific? – David Quinn Aug 25 '15 at 09:02
Perhaps you could explain what your interpretation of the question is, in particular the part about not using expansion? – Hypergeometricx Aug 25 '15 at 09:58

score 1 · Answer 3 · edited Apr 13 '17 at 12:20

1

You may use the following (discrete version of $\int x^k\,dx = \frac{x^{k+1}}{k+1}$):

Lemma: if $p(x)$ is a polynomial having degree $k$, then: $$ P(n) = \sum_{k=1}^{n}p(k) $$ is a polynomial with degree $k+1$.

That lemma gives that both sides are third degree polynomials in $n$. If you check that both sides agree for $n=1,2,3,4$, it follows that they are the same polynomial, hence your identity holds for every $n$.

Proof of the lemma: we may write any polynomial in the binomial base, then the well-known identity: $$ \sum_{k=1}^{n}\binom{k}{r} = \binom{n+1}{r+1} $$ proves the claim.

edited Apr 13 '17 at 12:20

Community

1

answered Aug 24 '15 at 18:03

Jack D'Aurizio

353,855

The statement "we may write any polynomial in the binomial base" would seem to require a proof, or, at least, a link. Personally, I have never seen the phrase "binomial base" though I understand what it means and can readily supply a proof. – marty cohen Aug 24 '15 at 18:37
1

@martycohen: well, we just need induction. $$x^k-k!\binom{x}{k}$$ is a polynomial having degree strictly less than $k$. – Jack D'Aurizio Aug 24 '15 at 19:11
That is a good way to do it. Maybe, while doing that, you could also throw in a proof that a polynomial takes only integer values at the integers iff it is the sum of integers times those binomials. – marty cohen Aug 25 '15 at 01:10
Thanks for your solution. I agree that using binomials for summation is a useful technique. However, to show that it is the sum of squares would require expansion of the binomials. The objective was to arrive at the answer using basic reindexing, etc. and without expansion. – Hypergeometricx Aug 25 '15 at 08:13
@hypergeometric: wait, I just used the fact that a decomposition exists to prove the claim. $i^2$ is a second-degree polynomial in $i$, hence $\sum_{i=1}^{n}i^2$ is a third-degree polynomial in $n$, without computing anything. – Jack D'Aurizio Aug 25 '15 at 13:07
Yes that's correct - point taken. So in your answer just after the yellow box are you saying that we would just have to check the LHS=RHS in the equation to be proven for $n=1,2,3,4$? – Hypergeometricx Aug 25 '15 at 15:15
@hypergeometric: exactly so. – Jack D'Aurizio Aug 25 '15 at 15:22
Thanks. This is quite an interesting approach! (+1) – Hypergeometricx Aug 25 '15 at 15:36

score 0 · Answer 4 · answered Apr 01 '16 at 11:15

$$\begin{align} \sum_{i=1}^n\sum_{j=1}^i2(n-i)+1 &=\sum_{i=1}^n\sum_{j=1}^i 2(n-i+1)-1\\ &=\sum_{s=1}^n\sum_{r=s}^n (2s-1) && (\text{putting }s=n-i+1\text{ and } r=n-j+1)\\ &=\sum_{r=1}^n\sum_{s=1}^r (2s-1) &&(1\le s\le r\le n)\\ &=\sum_{r=1}^n\sum_{s=1}^r s^2-(s-1)^2\\ &=\sum_{r=1}^n r^2\qquad\blacksquare \end{align}$$

Surprising Summation (3): $\sum_{i=1}^n\sum_{j=1}^i 2(n-i)+1=\sum_{i=1}^n i^2$

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