summation - Combination of quadratic and arithmetic series

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

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}