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Is there any way to find the sum of the below series ?

$$\underbrace{10^2 + 14^2 + 18^2 +\cdots}_{41\text{ terms}}$$

I got asked this question in a competitive exam.

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This series equals

$$\sum_{n=0}^{40} (10+4n)^2= 4\sum_{n=0}^{40} (5+2n)^2= 4\sum_{n=0}^{40} (25+20n+4n^2)= 4\left(25\sum_{n=0}^{40}1 +20\sum_{n=0}^{40}n+4\sum_{n=0}^{40}n^2\right)$$

Now recall that

$$\sum_{n=0}^m 1=m+1$$

$$\sum_{n=0}^m n=\frac{m(m+1)}{2}$$

$$\sum_{n=0}^m n^2=\frac{m(m+1)(2m+1)}{6}$$

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  • $\begingroup$ $ \begin{array}{l} D = \mathop {10^2 + 14^2 + 18^2 + ....}\limits_{41termes} \\ D = \sum\limits_{n = 0}^{40} {\left( {10 + 4n} \right)^2 = 4\sum\limits_{n = 0}^{40} {\left( 5 \right)^2 } + 4\sum\limits_{n = 0}^{40} {20n + 4} } \sum\limits_{n = 0}^{40} {4n^2 } \\ D = 4 \times 41 \times 25 + 80 \times \frac{{41}}{2} \times 800 + 16\frac{{40}}{6}\left( {41} \right)\left( {81} \right) \\ D = 4100 + 1312000 + 354240 = 1670340 \\ \end{array} $ $\endgroup$ – sabaga Jun 11 '14 at 19:59
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Let $$A=10^2+14^2+18^2+\dots$$ to 41 terms. Then $A=4B$, where $$B=5^2+7^2+9^2+\dots$$ to 41 terms. Let $$C=6^2+8^2+10^2+\dots$$ to 41 terms. Then $$B+C=5^2+6^2+7^2+\dots=(1^2+2^2+3^2+\dots)-(1^2+2^2+3^2+4^2)$$ where the sums go to $86^2$. Also, $C=4D$, where $$D=3^2+4^2+5^2+\dots=(1^2+2^2+3^2+\dots)-(1^2+2^2)$$ where the sums go to $43^2$. So you can calculate $D$ from Argon's last formula, then from $D$ you can get $C$; you can get $B+C$ from Argon, then $B$, then, finally, $A$.

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