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I encountered the following induction proof on a practice exam for calculus:

$$\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$$

I have to prove this statement with induction.

Can anyone please help me with this proof?


marked as duplicate by Arnaud D., dantopa, José Carlos Santos, J. W. Tanner, Jyrki Lahtonen Apr 16 at 2:50

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  • $\begingroup$ Well, is it true for $n=1$? If it is true for $n$, and you add $(n+1)^2$, does the formula still hold? $\endgroup$ – copper.hat Jul 3 '13 at 16:05

If $P(n): \sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6},$

we see $P(1): 1^2=1$ and $\frac{1(1+1)(2\cdot1+1)}{6}=1$ so, $P(1)$ is true

Let $P(m)$ is true, $$\sum_{k=1}^mk^2 = \frac{m(m+1)(2m+1)}{6}$$

For $P(m+1),$

$$ \frac{m(m+1)(2m+1)}{6}+(m+1)^2$$



$$=\frac{(m+1)(m+2)\{2(m+1)+1\}}6$$ as $m(2m+1)+6(m+1)=2m^2+7m+6=(m+2)(2m+3)$

So, $P(m+1)$ is true if $P(m)$ is true

  • $\begingroup$ I got the same until your very last step, thank you very much! $\endgroup$ – Nedellyzer Jul 3 '13 at 16:08
  • $\begingroup$ @SjoerdSmaal, my pleasure. $\endgroup$ – lab bhattacharjee Jul 3 '13 at 16:10
  • $\begingroup$ @labbhattacharjee: I think you should write it as: Let statement $P(n):\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$, then we can discuss whether $P(n)$ is true or not. The way it is now, $P(n)$ is an integer, not a statement, so it being 'true' doesn't make much sense. $\endgroup$ – Aang Jul 3 '13 at 16:14
  • $\begingroup$ @Avatar, do you have the latest version of the answer? $\endgroup$ – lab bhattacharjee Jul 3 '13 at 16:16
  • $\begingroup$ @labbhattacharjee: It's fine now :) $\endgroup$ – Aang Jul 3 '13 at 16:17

\begin{align} \sum_{k=1}^{n+1} k^2 & = \left(\sum_{k=1}^n k^2\right) & {} + (n+1)^2 \\[10pt] & = \underbrace{\left(\frac{n(n+1)(2n+1)}{6}\right)} & {} + (n+1)^2\tag{1} \end{align}

What you need is the same expression that you see over the $\underbrace{\text{underbrace}}$ but with $n+1$ in place of $n$. That would be $$ \frac{[n+1]\Big([n+1]+1\Big)\Big(2[n+1]+1\Big)}{6}.\tag{2} $$

So the problem is to show that $(1)$ is equal to $(2)$. If you can be more explicit about where you ran into difficulties, I could possibly say more.


HINT: $$\sum_{k=1}^{n+1}k^2=\sum_{k=1}^nk^2+(n+1)^2=\frac{n(n+1)(2n+1)}{6}+(n+1)^2$$

Now simplify and show that it is equivalent to replacing $n$ by $n+1$ in the original formula.

  • $\begingroup$ Thanks you, I already got this but couldn't manage to write down the last step, I know how to do it now! $\endgroup$ – Nedellyzer Jul 3 '13 at 16:09

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