Combination of quadratic and arithmetic series

Question

Problem:

Calculate $\dfrac{1^2+2^2+3^2+4^2+\cdots+23333330^2}{1+2+3+4+\cdots+23333330}$.

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Attempt:

I know the denominator is arithmetic series and equals $$\frac{n}{2}(T_1+T_n)=\frac{23333330}{2}(1+23333330)=272222156111115,$$ but how do I calculate the numerator without using a calculator?

Answer 1

Intuitively, \begin{align} S_1&=\frac{1^2}{1}=1=\frac{3}{3}\\ S_2&=\frac{1^2+2^2}{1+2}=\frac{5}{3}\\ S_3&=\frac{1^2+2^2+3^2}{1+2+3}=\frac{7}{3}\\ S_4&=\frac{1^2+2^2+3^2+4^2}{1+2+3+4}=3=\frac{9}{3}\\ \vdots\\ \large\color{blue}{S_n}&\color{blue}{=\frac{2n+1}{3}}. \end{align}

Answer 2

$$S_1=\sum_{r=1}^nr =\frac{n(n+1)}2$$

and $$S_2=\sum_{r=1}^nr^2=\frac{n(n+1)(2n+1)}6$$

So, the ratio $\dfrac{S_2}{S_1}$ should not demand calculator