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25. Find the sum of the following series up to n terms: $\frac{1^3}{1} + \frac{1^3+2^3}{1+3}+ \frac{1^3+2^3+3^3}{1+3+5}+ ...$

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n term of series :

$\frac{1^3}{1} + \frac{1^3+2^3}{1+3}+ \frac{1^3+2^3+3^3}{1+3+5}+ ........=\frac{1^3+2^3+3^3+..........n^3}{1+3+5+...........(2n-1)}$

$=\frac{\left [ \frac{n(n+1)}{2} \right ]^2}{1+3+5+............(2n-1)}$

Here, $1,3,5............(2n-1)$ are in AP with first term =a=1 , last term = 2n-1, number of terms =n

$1+3+5............(2n-1)=\frac{n}{2}\left [ 2(1)+(n-1)2 \right ]$

$=\frac{n}{2}\left [ 2+2n-2 \right ]=n^2$

$a_n=\frac{n^2(n+1)^2}{4n^2}$

$=\frac{(n+1)^2}{4}$

$=\frac{n^2}{4}+\frac{n}{2}+\frac{1}{4}$

$S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} \frac{(k+1)^2}{4}$

$=\frac{1}{4}\sum _{k=1}^{n} k^2+\frac{1}{2}\sum _{k=1}^{n} k+\sum _{k=1}^{n}\frac{1}{4}$

$=\frac{1}{4}\frac{n(n+1)(2n+1)}{6}+\frac{1}{2}\frac{n(n+1)}{2}+\frac{n}{4}$

$=n\left ( \frac{(n+1)(2n+1)}{24}+\frac{n+1}{4}+\frac{1}{4} \right )$

$=n\left ( \frac{(n+1)(2n+1)+6(n+1)+6}{24} \right )$

$=n\left ( \frac{2n^2+n+2n+1+6n+6+6}{24} \right )$

$=n\left ( \frac{2n^2+9n+13}{24} \right )$

Posted by

seema garhwal

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