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I was trying to show that $\sum \limits_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ but instead I got this $[\frac{n(n+1)}{2}]^2$ which from my understanding I basically proved another summation formula which is $\sum \limits_{k=1}^n k^3$. Obviously I must have done something wrong. So I am going to show you how I got the summation formula wrong.

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

In reverse order

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

I decided to square root the partial sum which probably what led to the wrong answer. But I do not know, I am clumsy when I write on paper.

$\sqrt{s_n} = 1 + 2 + ... + (n-1) + n$

In reverse order

$\sqrt{s_n} = n + (n-1) + ... + 2 + 1$

Adding the two partial sums

$\sqrt{s_n} + \sqrt{s_n} = 2\sqrt{s_n}$

$2\sqrt{s_n} = (n+1) + (n+1) + ...$

$2\sqrt{s_n} = n(n+1)$

$\sqrt{s_n} = \frac{n(n+1)}{2}$

$(\sqrt{s_n})^2 = [\frac{n(n+1)}{2}]^2$

Now my question what did I do wrong. Can somebody show me the correct way. I am pretty sure this a fake proof, or a minor error. Thank you.

P.S. I am no latex expert and this not homework just for practice.

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2 Answers 2

up vote 6 down vote accepted

$\sqrt{a^2+b^2}\ne a+b$ in general unless at least one of $a,b$ is $0$

If $s_n=1^2+2^n+\cdots+(n-1)^2+n^2,$

how can you write $s_n=1+2+\cdots+(n-1)+n?$

(1)One way to proof is :

$ (r+1)^3-r^3=3r^2+3r+1$

Put $r=0,1,2,\cdots,n-1,n$ and add to get

$(n+1)^3=3S_n+3(1+2+3+\cdots+n)+n=3S_n+3\frac{n(n+1)}2+n$

So, $S_n=...$

(2)We can use induction too:

Let $S(m)= \frac{m(m+1)(2m+1)}6$

$S(1)=\frac{1\cdot\cdot3}6=1$ which is true.

So, $S(m+1)=S(m)+m+1$ $=\frac{m(m+1)(2m+1)}6+(m+1)^2=\frac{(m+1)\{(m+1)+1\}\{2(m+1)+1\}}6$

So the proposition holds true for $n=m+1$ if it is true for $n=m.$

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Oh you are right. I total forgot about. Thanks. So can you show the proof of k^2 please? –  Daniel Lopez Dec 24 '12 at 20:03
    
Thanks now I get I will mark your answer. Great explanation. –  Daniel Lopez Dec 24 '12 at 20:13

$S^2_n= 1^2+2^2+3^2+...+n^2$

$Let, S_n=1+2+3+...+n$

$(S_n)^2=(1^2+2^2+3^2+...+n^2)+\left((1*2+1*3+...+1*n)+(2*1+2*3+...+2*n)+...+(n*1+1*2+n*3+...+n*(n-1))\right)$ $(S_n)^2=S^2_n+\left((1*2+1*3+...+1*n)+(2*1+2*3+...+2*n)+...+(n*1+1*2+n*3+...+n*(n-1))\right)$ $\sqrt[]{(S_n)^2}=\sqrt[]{S^2_n+\left((1*2+1*3+...+1*n)+(2*1+2*3+...+2*n)+...+(n*1+1*2+n*3+...+n*(n-1))\right)}$

$S_n=\sqrt[]{S^2_n+\left((1*2+1*3+...+1*n)+(2*1+2*3+...+2*n)+...+(n*1+1*2+n*3+...+n*(n-1))\right)}$

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What? Please define what you are summing... The first line is wrong. –  anorton Dec 25 '12 at 1:47
    
$S^2_n$ is a notation which says the sum of the squares of the first n natural numbers. –  Rajesh K Singh Dec 25 '12 at 1:54
    
Similarly, $S_n$ stands for the sum of the first n natural numbers. –  Rajesh K Singh Dec 25 '12 at 1:55
    
But $\sqrt{S_n^2} \not = S_n$. –  anorton Dec 25 '12 at 1:59
1  
Ok. I see you've edited your answer. –  anorton Dec 25 '12 at 14:25

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