Prove a formula is corect I am trying to figure out this discrete math problem. I am not sure how to do it or even how to really start it. The problem is as follows: 
Consider values of $\frac{\sum_{i=1}^n i^2}{\sum_{i=1}^n i}$ for several small values of n. What formula will express $\sum_{i=1}^n i^2$ in terms of $n$. Prove this is a correct formula. 
 A: Let $$f(n)=\frac{\sum_{i=1}^ni^2}{\sum_{i=1}^ni}\;.$$ Begin by following the instructions: compute $f(n)$ for some small values of $n$.
$$\begin{array}{r|cc}
n&1&2&3&4&5&6&7\\
\sum_{i=1}^ni^2&1&5&14&30&55&91&140\\
\sum_{i=1}^ni&1&3&6&10&15&21&28\\
f(n)&1&\frac53&\frac73&3&\frac{11}3&\frac{13}3&5
\end{array}$$
That last line showing $f(n)$ looks awfully regular. If the pattern isn’t immediately apparent, try putting everything over a denominator of $3$: $\frac33,\frac53,\frac73,\frac93,\frac{11}3,\frac{13}3,\frac{15}3$. Those numerators are clearly $2n+1$, at least as far as this table goes, so we conjecture that $$f(n)=\frac{2n+1}3$$ for all positive integers $n$.
If this conjecture is correct, $$\sum_{i=1}^ni^2=f(n)\sum_{i=1}^ni=\frac13(2n+1)\sum_{i=1}^ni\;.\tag{1}$$
I expect that you already know that $\sum_{i=1}^ni=\frac{n(n+1)}2$. (If not, it follows immediately from the formula for the sum of an arithmetic progression.) Combine this with $(1)$ to get a formula for $\sum_{i=1}^ni^2$ that doesn’t involve any summations. Then use mathematical induction to prove that your formula is correct.
A: 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 $\sum_{i=1}^n i^2$. 
