You are asked to calculate. So let us calculate. Let $f(n)=\frac{\sum_{i=1}^n i^2}{\sum_{i=1}^{n} i}.$ We have $f(1)=\frac{1^2}{1}=1,\quad f(2)=\frac{5}{3},\quad f(3)=\frac{14}{6}=\frac{7}{3},\quad f(4)=\frac{30}{10}=3,\quad f(5)=\frac{11}{3}.$
Maybe make all the denominators equal to $3$. We get $\dfrac{3}{3}$, $\dfrac{5}{3}$, $\dfrac{7}{3}$, $\dfrac{9}{3}$, $\dfrac{11}{3}$. Nice pattern!
We might conjecture on the basis of the evidence so far that $f(n)=\dfrac{2n+1}{3}$. Calculation of the next few terms seems to confirm that.
Now for a proof. Your post hints that you might know a simple formula for $\sum_{i=1}^n i$, and you want a formula for $\sum_{i=1}^n i^2$. There is such a formula, it is $\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}.$ The induction proof is not very complicated.
Remark: Here is a nice way of finding $\sum_{i=1}^n i^2$. Note that $i^3-(i-1)^3=3i^2-3i+1.$ Add up, $i=1$ to $n$. On the left, there is almost total cancellation (telescoping) and we get $n^3$. It follows that $n^3=3\sum_{i=1}^n i^2-3\sum_{i=1}^n i +\sum_{i=1}^n 1.$ Now from the fact that $\sum_{i=1}^n i=\dfrac{n(n+1)}{2}$ and some algebra we get the formula for $\sum_{i=1}^n i^2$.