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I have $1^2+2^2+...+n^2$ and I want to prove the sum is $\frac{m(m+1)(2m+1)}{6}$. So for proving the formula for $1+3+5+7+... = n^2$ this is how i got the formula:

the common difference is 2, so the formula for $a_n = 2n-1$

The sum of an arithmetic series is $\frac{n(a_{1}+a_{n})}{2}$ so I have $\frac{n(1+(2n-1)}{2}$ so its $n^2$.

For the sum of the squares, I know that the term for $a_n$ is $n^2$. But since it's not an arithmetic sequence, I can't use my previous formula. Is there an analogous method?

I already know how to prove this by induction and i understand the visual proofs. But if you see how I formulated my sum for the first consecutive odd integers, can you please show me how I can do this for the sum of the squares in the same way or just tell me it's not possible if it isn't. Thank you!

user8290579
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  • Not sure if this is what you're asking for - but one possibility is $\sum_{k=1}^n k^2 = \sum_{k=1}^n \left( 2\binom{k}{2} + k \right) = 2 \binom{n+1}{3} + \binom{n+1}{2}$. – Daniel Schepler Oct 31 '18 at 19:09
  • thanks! that is what I meant in terms of finding another method but I didn't want to use binomial coefficient. basically im trying to explain this to a kid in algebra II who just learned sequences and series so he doesn't know how to compute that – user8290579 Oct 31 '18 at 19:12

3 Answers3

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hint

Your approach $$1^2=1$$ $$2^2=1+3$$ $$3^2=1+3+5$$ $$4^2=1+3+5+7$$

$$n^2=1+3+5+...+(2n-1)$$

thus

$$1^2+2^2+...+n^2=n+3(n-1)+5(n-2)+...+2(2n-3)+(2n-1)$$

it complicate things.

an other approach

$$(n+1)^3=n^3+3n^2+3n+1$$ $$n^3=(n-1)^3+3(n-1)^2+3(n-1)+1$$ $$(n-1)^3=(n-2)^3+3(n-2)^2+3(n-2)+1$$ ... $$2^3=1^3+3+3+1$$ $$1^3=1$$

by sum $$(n+1)^3=3(1^1+2^2+...+n^2)+3(1+2+...+n)+(n+1)$$

the result is $$\frac{(n+1)((n+1)^2-1-\frac 32 n)}{3}$$

hamam_Abdallah
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It comes down to whether you can prove $\sum_{i=0}^n\binom{i}{j}=\binom{n+1}{j+1}$ without induction. Since $i^2=2\binom{i}{2}+i$, $$\sum_{i=0}^n i^2=2\binom{n+1}{3}+\binom{n+1}{2}=\frac{n(n+1)}{6}(2(n-1)+3)=\frac{n(n+1)(2n+1)}{6}.$$So let's try a combinatorial proof of that sum in binomial coefficients (an inductive proof is trivial). The right-hand side is the $x^{j+1}$ coefficient in $(1+x)^{n+1}$. Numbering the $n+1$ linear factors from $0$ to $n$ inclusive, if the final coefficient to contribute an $x$ factor towards an $x^{j+1}$ term is factor $i$, we must choose $x$ from $j$ of the first $i$ factors (i.e. those numbered from $0$ to $i-1$).

J.G.
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    Or, split the $j+1$-element subsets of ${ 1, 2, \ldots, n+1 }$ according to their maximum element: there are $\binom{k-1}{j}$ such subsets with maximum element $k$. – Daniel Schepler Oct 31 '18 at 21:18
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Not really. The reason you get a nice easy formula for adding arithmetic series is distributivity.

Notice $\sum_{i=1}^{n}a_i$ where $a_i=mi+k$ is just $$\sum_{i=1}^n(mi+k)=(\sum_{i=1}^nmi)+(\sum_{i=1}^nk)=m(\sum_{i=1}^ni)+nk=m\frac{n(n+1)}{2}+nk=n\frac{(m(n+1)+2k)}{2}=n\frac{(mn+k+m+k)}{2}=\frac{n(a_1+a_k)}{2}$$

The reason this works is because the terms of an arithmetic series are extremely basic and so you can use commutativity and distributivity to reduce the problem to just summing the first $n$ integers and playing around with that. When you are trying to add squares there is no easy way to factor out so that you get to such a basic form. Notice that even here the hard part really was the one we left out and that's adding the first $n$ integers.

DRF
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