Question

If the sum of the $3^3+7^3+{11}^3+{15}^3+...\text{upto}20$ terms is $S_{20}$. Then the value of $S_{20}$ is___

Answer 1

Let $S=3^3+7^3+{11}^3+{15}^3+....$

Above given series is sum of the cube of odd numbers starting from $3$ with common difference $4.$

$n\text{th}$ term of above given series is
$t_n={[3+(n-1)\times 4]}^3=(4n-1{)}^3$

$S_n=\sum _{i=1}^n{t}_n$
$=\sum _{i=1}^n$ $(4n-1{)}^3$
$=\sum _{i=1}^n[64n^3-48n^2+12n-1]$
$=64\sum _{i=1}^n{n}^3-48\sum _{i=1}^n{n}^2+12\sum _{i=1}^{n}n-\sum _{i=1}^{n}1$
$=64{\left[\frac{n(n+1)}{2}\right]}^2-48\left[\frac{n(n+1)(2n+1)}{6}\right]+12\left[\frac{n(n+1)}{2}\right]-n$

Substitute $n=20$ in the above equation

$S_{20}=64{\left[\frac{(20\times 21)}{2}\right]}^2-48\left[\frac{(20\times 21\times 41)}{6}\right]+12\left[\frac{(20\times 21)}{2}\right]-20$

$=2822400-137760+2520-20$

$S_{20}$ = 2687140