Question

Find sum of series upto n terms -

$\dfrac{ {1}^{3} }{1} + \dfrac{ {1}^{3} + {3}^{3} }{1 + 3} + \dfrac{ {1}^{3} + {3}^{3} + {5}^{3} }{1 + 3 + 5} + \dots$

Answer 1

ANSWER:

To Find:

$\text{Value of:} \dfrac{1^3}{1}+\dfrac{1^3+3^3}{1+3}+\dfrac{1^3+3^3+5^3}{1+3+5}+\dots n\:terms$

Solution:

We need to find the value of,

$\implies \dfrac{1^3}{1}+\dfrac{1^3+3^3}{1+3}+\dfrac{1^3+3^3+5^3}{1+3+5}+\dots n\:terms$

The $\sf k^{th}$term will be:

$\implies T_{k}=\dfrac{1^3+3^3+5^3+\dots+k^3}{1+3+5+\dots +k}$

We know that,

$\hookrightarrow 1^3+3^3+5^3+\dots+n^3=n^2(2n^2-1)$

$\hookrightarrow 1+3+5+\dots+n= n^2$

So,

$\implies\sf T_{k}=\dfrac{1^3+3^3+5^3+\dots+ k^3}{1+3+5+\dots+k}$

$\implies \sf T_{k}=\dfrac{k^2\!\!\!\!/\:(2k^2-1)}{k^2\!\!\!\!/}$

$\implies \sf T_{k}=2k^2-1$

Now, we have our $\sf k^{th}$ term.

So, the required sum is,

$\displaystyle\implies \sf \sum\limits_{k=1}^{n} T_k$

$\displaystyle\implies \sf \sum\limits_{k=1}^{n} (2k^2-1)$

$\displaystyle\implies \sf 2\sum\limits_{k=1}^{n}k^2- \sum\limits_{k=1}^{n} 1$

We know that,

$\hookrightarrow \sf \sum\limits_{k=1}^{n}k^2=\dfrac{n(n+1)(2n+1)}{6}$

$\hookrightarrow \sf \sum\limits_{k=1}^{n}1=n$

$\displaystyle\implies \sf 2\sum\limits_{k=1}^{n}k^2- \sum\limits_{k=1}^{n} 1$

$\displaystyle\implies \sf 2\!\!\!/\:\dfrac{n(n+1)(2n+1)}{6\!\!\!/\:_{3}}-n$

$\displaystyle\implies \sf \dfrac{n(n+1)(2n+1)}{3}-n$

$\displaystyle\implies \sf \dfrac{n(n+1)(2n+1)-3n}{3}$

$\displaystyle\implies \sf \dfrac{(n^2+n)(2n+1)-3n}{3}$

$\displaystyle\implies \sf \dfrac{2n^3+n^2+2n^2+n-3n}{3}$

$\displaystyle\implies \sf \dfrac{2n^3+3n^2-2n}{3}$

$\displaystyle\implies \sf \dfrac{n(2n^2+3n-2)}{3}$

$\displaystyle\implies \sf \dfrac{n(2n^2+4n-n-2)}{3}$

$\displaystyle\implies \sf \dfrac{n[2n(n+2)-1(n+2)]}{3}$

$\displaystyle\implies \sf \dfrac{n(2n-1)(n+2)}{3}$

So,

$\displaystyle\implies \sf \dfrac{n(n+2)(2n-1)}{3}$

Hence, the required sum is:

$\displaystyle\implies \bf \dfrac{n(n+2)(2n-1)}{3}$