Find the sum of the following series up to n terms:


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


The $n^{\text {th }}$ term of the given series is $\frac{1^{3}+2^{3}+3^{3}+\ldots+n^{3}}{1+3+5+\ldots+(2 n-1)}=\frac{\left[\frac{n(n+1)}{2}\right]^{2}}{1+3+5+\ldots+(2 n-1)}$

Here, $1,3,5, \ldots(2 n-1)$ is an A.P. with first term a, last term $(2 n-1)$ and number of terms as $n$

$\therefore 1+3+5+\ldots .+(2 n-1)=\frac{n}{2}[2 \times 1+(n-1) 2]=n^{2}$

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

$\therefore \mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{K}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{K}}=\sum_{\mathrm{K}=1}^{\mathrm{n}}\left(\frac{1}{4} \mathrm{~K}^{2}+\frac{1}{2} \mathrm{~K}+\frac{1}{4}\right)$

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

$=\frac{n[(n+1)(2 n+1)+6(n+1)+6]}{24}$

$=\frac{n\left[2 n^{2}+3 n+1+6 n+6+6\right]}{24}$

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

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