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This question already has an answer here:

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?

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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
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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 $P(m+1),$

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

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

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

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

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  • $\begingroup$ 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
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\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.

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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.

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  • $\begingroup$ 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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