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I want to prove by induction that,

$\sum\limits_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$

OK I got the initial step, however, I have problems with the induction step:

Here is what I tried:

$\sum\limits_{i=1}^{n+1} i^2 + n+1= \frac{n(n+1)(2n+1)}{6} + n+1 =\frac{n(n+1)(2n+1)}{6} + \frac{6*(n+1)}{6} = \frac{(n+1)(n+6)(2n+1)}{6}$

Here I am stuck. How to conclude this for $A(n+1)$?

I appreciate your answers!

UPDATE:

$\sum\limits_{i=1}^{n+1} i^2 + n+1= \frac{n(n+1)(2n+1)}{6} + n+1 =\frac{n(n+1)(2n+1)}{6} + \frac{6*(n+1)}{6} = \frac{(n+1)*[n(2n+1) + 6]}{6}$

Martin Sleziak
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Carol.Kar
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2 Answers2

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$\sum_{i=1}^{n+1}i^{2}=\sum_{i=1}^{n}i^{2}+(n+1)^{2}$

by the induction hypothesis we have

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

$=\frac{2n^{3}+3n^{2}+n}{6}+\frac{6n^{2}+12n+6}{6}=\frac{2n^{3}+9n^{2}+13n+6}{6}=\frac{(2n^{3}+9n^{2}+7n)+(6n+6)}{6}=\frac{n(2n+7)(n+1)+6(n+1)}{6}=\frac{(n+1)(2n^{2}+7n+6)}{6}$

$=\frac{(n+1)(2n+3)(n+2)}{6}=\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$.

user71352
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If the induction is to work we need $\sum_{i=1}^{n+1}{i^2}$ = (n+1)(n+2)(2n+3) which we get by substituting (n+1) for n in the proposed formula. We have

$\sum_{i=1}^{n+1}{i^2}$ = $\sum_{i=1}^{n}{i^2}$ + $(n+1)^2$ = $\frac{n(n+1)(2n+1)}{6}$ + $(n+1)^2$ = $\frac{n(n+1)(2n+1) + 6(n+1)^2}{6}$ . We can now deal just with the numerator and we have

n(n+1)(2n+1) + 6$(n+1)^2$ = (n+1)[n(2n+1) + 6(n+1)] = (n+1)[2$n^2$ + 7n + 6] = (n+1)(n+2)(2n+3).

That is exactly what the numerator should be, so the induction step is done.

Betty Mock
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