# Combination of quadratic and arithmetic series

Problem:

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

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?

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Do you know how to find the sum of squares? –  Calvin Lin Jul 12 '14 at 19:34
I dunno. Can you help me Calvin? –  L Lawliet Jul 12 '14 at 19:36
Lab stated the formulas in his solution. –  Calvin Lin Jul 12 '14 at 19:37

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}

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This solution appears to be more elegant than others! –  MonK Jul 12 '14 at 20:09
This is smart way. Genius! (y)(y) –  L Lawliet Jul 12 '14 at 20:10

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

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

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What do you think the OP learned from seeing this? One more formula to be memorized? –  Jyrki Lahtonen Jul 12 '14 at 19:38
@doank, See : math.stackexchange.com/questions/183316/… –  lab bhattacharjee Jul 12 '14 at 19:41
How to get the formula sum of squares? –  L Lawliet Jul 12 '14 at 19:42
@doank: when search $S_2(n)$ as polynomial $S_2(n)=an^3+bn^2+cn$ (due to intuition :), then $$S_2(n)-S_2(n-1) = n^2$$ can help here. –  Oleg567 Jul 12 '14 at 19:46
@doank,$$S_2(n)= 1^2+2^2+\ldots+(n-1)^2+n^2,$$ $$S_2(n-1)= 1^2+2^2+\ldots+(n-1)^2,$$ so $$S_2(n) = S_2(n-1)+n^2,\tag{1}$$ or $$S_2(n)-S_2(n-1) = n^2,\tag{2}$$ Now, if you'll search $S(n)$ as $$S(n)=an^3+bn^2+cn,$$ ($S(0)=0$, so last term "$+d$" is void), then $(2) \Rightarrow$ $$an^3+bn^2+cn - a(n-1)^3 - b(n-1)^2-c(n-1)=n^2;$$ $$an^3+bn^2+cn - a(n^3-3n^2+3n-1) -b(n^2-2n+1)-c(n-1)=n^2;$$ $$3a n^2 +(-3a+2b)n + (a-b+c)=n^2;$$ so $$3a=1;\\ -3a+2b=0;\\ a-b+c=0;$$ and hence we can find $a=\dfrac{1}{3}$, $b=\dfrac{1}{2}$, $c=\dfrac{1}{6}$. –  Oleg567 Jul 12 '14 at 20:48