Question

If the sum of the $3^3+7^3+{11}^3+{15}^3...................20$ terms is $S_{20}$. Find the value of $\frac{S_{20}}{100}$ .

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Answer 1

Let S = $3^3+7^3+{11}^3+{15}^3+.........\left(1\right)$

Above given series is sum of the cube of odd numbers starting from 3.

$n_{th}$ term of above given series

$t_n$ = ${[3+(n-1)\times 4]}^3$ = ${[3+4n-4]}^3$ = $(4n-1{)}^3$

sum of the n terms

$S_n={\sum }_{i=1}^n{t}_n$

= ${\sum }_{i=1}^n$ $(4n-1{)}^3$

= ${\sum }_{i=1}^n$ $[16n^3-1-4n(4n-1\left)\right]$

= ${\sum }_{i=1}^n$ $[16n^3-1-16n^2+4n]$

= 16 ${\sum }_{i=1}^n{n}^3$ - ${\sum }_{i=1}^n$1 - 16${\sum }_{i=1}^n{n}^2$ + 4${\sum }_{i=1}^{n}n$

= $16{\left[\frac{n(n+1)}{2}\right]}^2$ - n - $16\left[\frac{n(n+1)(2n+1)}{2}\right]$ + 4 [$\frac{n(n+1)}{2}$] ---------(2)

substitute n = 20 in equation (2)

$=16{\left[\frac{(20\times 21)}{2}\right]}^2$ - 20 - $16\left[\frac{(20\times 21\times 41)}{2}\right]$ + 4 [$\frac{(20\times 21)}{2}$]

$=705600-2045920+840$

$S_{20}=660500$

$\frac{S_{20}}{100}=6605$