Prove $1^2+2^2+3^2+4^2...+n^2=\dfrac{n(n+1)(2n+1)}6$ by induction.
I am trying to show true for $n=k+1$.
I now have the equation $1^2+2^2+3^2+4^2...+k^2+(k+1)^2=\dfrac{k(k+1)(2k+1)}6+(k+1)^2$. I know that the result should be $1^2+2^2+3^2+4^2...+k^2+(k+1)^2=\dfrac{(k+1)(k+2)(2k+3)}6$ How can I achieve this?
Its good you know that $$1^2+\cdots +(k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$$.
Taking the lcm, we obtain that $$1^2+\cdots +(k+1)^2 = \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6}.$$
By factorization, you get $$1^2+\cdots +(k+1)^2 = \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6} = \frac{(k+1)[k(2k+1)+6(k+1)]}{6}$$
= $$ \frac{(k+1)(2k^2+k+6k+6)}{6}= \frac{(k+1)(2k^2+7k+6)}{6}$$
I will let you factorise $2k^2+7k+6$ yourself to get your desired answer