induction with recursive elements $y_{k}=y_{k-1}+k^2$

Question

the problem is $y_{k}=y_{k-1}+k^2$, for all integers k $>=$ 2 given $y_{1}=1$

Honestly I got caught up with this question

$y_{2}=1+2^2 = 1+4 = 5$

$y_{k+1}=y_{k}+(k+1)^2$ -plugging in $y_{k}$

$y_{k+1}=y_{k-1}+k^2+(k+1)^2$ - foiling out $(k+1)^2$

$y_{k+1}=y_{k-1}+k^2+k^2+2k+1$

I have no clue where to go from here or if i'm even on the right track if anyone can give me some help it would be appreciated thank you

Answer 1

We have $y_{k}=y_{k-1}+k^2$, so \begin{eqnarray*} y_{k}=k^2+(k-1)^2+ \cdots+ 1 =\sum_{i=1}^{k} i^2 =\frac{n(n+1)(2n+1)}{6}. \end{eqnarray*}

Answer 2

In general, in any equation of the form $$y_k-y_{k-1}=f(k)$$ you can sum on both sides to get $$y_n-y_0=\sum_{k=1}^n f(k),$$ Via telescopic property at the lhs.