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:
$\sum\limits_{i=1}^{n+1} i^2 + n+1= \frac{n(n+1)(2n+1)}{6} + n+1 =\frac{n(n+1)(2n+1)}{6} + \frac{6*(n+1)}{6} = \frac{(n+1)*[n(2n+1) + 6]}{6}$
Best Answer
$\sum_{i=1}^{n+1}i^{2}=\sum_{i=1}^{n}i^{2}+(n+1)^{2}$
by the induction hypothesis we have
$=\frac{n(n+1)(2n+1)}{6}+(n+1)^{2}=\frac{(n^{2}+n)(2n+1)}{6}+\frac{6(n^{2}+2n+1)}{6}$
$=\frac{2n^{3}+3n^{2}+n}{6}+\frac{6n^{2}+12n+6}{6}=\frac{2n^{3}+9n^{2}+13n+6}{6}=\frac{(2n^{3}+9n^{2}+7n)+(6n+6)}{6}=\frac{n(2n+7)(n+1)+6(n+1)}{6}=\frac{(n+1)(2n^{2}+7n+6)}{6}$
$=\frac{(n+1)(2n+3)(n+2)}{6}=\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$.