Alternating sign square sum

Question

It can be seen that $6^2 - 5^2 + 4^2 - 3^2 + 2^2 - 1^2 =21 = T_6$, the sixth triangle number. Prove that the $n$ th triangle number, $T_n = n^2 - (n - 1)^2 + (n - 2)^2- \cdots + 1^2$.

Well my solution is :

if $n$, and group them into $n/2$ pairs:

\begin{align} S & = [n^2 - (n - 1)]^2 + \cdots + (6^2 - 5^2) + (4^2 - 3^2) + (2^2 - 1^2)\\ &= 2n - 1 + \cdots + 11 + 7 + 3\\ &= 3 + \cdots + 2n - 5 + 2n - 1\\ 2S&= \frac {n}{2}(2n + 2)\\ &= n(n + 1)\\ S &= \frac{n(n + 1)}{2} \end{align}

This is where I stopped. I need explanations please

I can't say who downvoted but if you don't show some self effort and self work in your own question I'm afraid downvotes will continue to arrive. People here expect askers to show they already tried something and show that in their questions. — DonAntonio, Feb 28 '14 at 14:43
Oh so that's the problem ok i already know the answer so i thought to that i can get more explanations — Manal, Feb 28 '14 at 14:45
Nice: then do write down your solution and either ask for explanation for it or for a new solution... — DonAntonio, Feb 28 '14 at 14:46
@Manal Here is an upvote (+1) because I like the problem. Now let's see some work here. — imranfat, Feb 28 '14 at 14:58

Ross Millikan · Answer 1 · 2014-03-01T15:34:22.960

You are trying to prove that for even $n$, $S(n)=\sum_{i=1}^n (-1)^ii^2=\frac 12n(n+1)=T(n)$ The base case is easy: $S(2)=2^2-1^2=3=T(2)$. Now assume it is true for $k$ and see if we can prove it for $k+2$$$\begin{align} S(k+2)&=\sum_{i=1}^{k+2} (-1)^ii^2\\ &=(k+2)^2-(k+1)^2+S(k)\\ &=2k+3+\frac 12k(k+1)\\ &=\frac 12(k+2)(k+3)\\&=T(k+2)\end{align}$$ where the second line comes by separating out the new terms and the third line used the induction hypothesis

Added: the proof can be direct. Let $k=2m$ The approach you are looking for goes like this: $$\sum_{i=1}^m ((2i)^2-(2i-1)^2)=\sum_{i=1}^m(4i-1)=4\cdot \frac 12m(m+1)-m=\frac 12(2m)(2m+1)=T(2m)$$

Another approach comes is inspired by $6^2-5^2+4^3-3^2+2^2-1^2=2(6^2+4^2+2^2)-(6^2+5^2+4^3+3^2+2^2+1^2)=8(3^2+2^2+1^2)-(6^2+5^2+4^3+3^2+2^2+1^2)$ and uses the sum of squares formula: $$S(2m)=8\sum_{i=1}^mi^2-\sum_{j=1}^{2m}j^2=8\cdot \frac 16m(m+1)(2m+1)-\frac 16(2m)(2m+1)(4m+1)=T(2m)$$

score 1 · Answer 2 · answered Feb 28 '14 at 15:08

1

Hint: What happens if you factor $n^2-(n-1)^2$?

answered Feb 28 '14 at 15:08

Mike

13,318

mm.. i think this: ($n - $n - $1)($n - $n - $1) – Manal Feb 28 '14 at 15:11
@Manal All the Tex goes between 2 dollar signs, don't just put a dollar sign in front of each variable. And even after fixing what is probably a typo, that's not quite right. – Mike Feb 28 '14 at 15:14
okay thanks for your advice – Manal Feb 28 '14 at 15:19

Alternating sign square sum

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