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$1^2 = 1$

$1^2 + 2^2 = 5$

$1^2 + 2^2 + 3^2 = 14$

$1^2 + 2^2 + 3^2 + 4^2 = 30$

$1^2 + 2^2 + 3^2 + 4^2 + ...... + n^2 = an^3 + bn^2 + (n/6)$

Work out the values of $a$ and $b$.

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$$\sum_{r=0}^nr^2=\frac{n(n+1)(2n+1)}{6}$$ now you can expand this

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  • $\begingroup$ just a question. $\endgroup$ – user547075 Nov 13 '18 at 3:00
  • $\begingroup$ shouldn't r start with 1? $\endgroup$ – user547075 Nov 13 '18 at 3:00
  • $\begingroup$ Possibly yes, although in this context the $r=0$ term is ignored obviously $\endgroup$ – Henry Lee Nov 13 '18 at 3:04
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Sum of squares of first $n$ natural numbers is given by : $$ \Sigma_{r=1}^nr^2= \frac{n(n+1)(2n+1)}{6}$$ Expanding it: $$\frac13 n^3+\frac 12 n^2+\frac n6$$

So, $$a=\frac 13$$ and $$b=\frac 12$$

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  • $\begingroup$ Thanks for the reply, but is it possible to get the values of a and b without using sigma? $\endgroup$ – V11 Nov 13 '18 at 4:40
  • $\begingroup$ We used sigma to get that expression. Once got that, then just simplify it to form a cubic. And you are done! $\endgroup$ – idea Nov 13 '18 at 5:27
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By one of the properties of sigma, the last equation can be expressed as $\frac{n(n+1)(2n+1)}{6}$.

This means that the sum of the squares of numbers from $1$ to $n$ can be expressed as $\frac{n(n+1)(2n+1)}{6}$.

Therefore, you just have to make it equal to that and work out the values of $a$ and $b$ by yourself.

To be more clear, you just need to let $an^3+bn^2+\frac{n}{6}= \frac{n(n+1)(2n+1)}{6}$ and solve for $n$.

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  • $\begingroup$ What do you mean exactly? $\endgroup$ – V11 Nov 13 '18 at 3:42
  • $\begingroup$ I edited my answer $\endgroup$ – user547075 Nov 13 '18 at 4:20
  • $\begingroup$ Thanks for the reply. I still have a few doubts though. How will solving for n get the values for a and b? And is it possible to get the values for a and b without using sigma? $\endgroup$ – V11 Nov 13 '18 at 4:44
  • $\begingroup$ 1^2+2^2+.....+n^2=an^3+bn^2+n/6=n(n+1)(2n+1)6 Is it better to understand?In case you are curious about how the formula is made, you can copy and go to this link. mathforum.org/library/drmath/view/56920.html . This shows you how the formula is achieved. $\endgroup$ – user547075 Nov 13 '18 at 21:57

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