Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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


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 $n=m.$

share|cite|improve this answer
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

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

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

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


share|cite|improve this answer
What? Please define what you are summing... The first line is wrong. – apnorton 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 $\sqrt{S_n^2} \not = S_n$. – apnorton Dec 25 '12 at 1:59
Ok. I see you've edited your answer. – apnorton Dec 25 '12 at 14:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.