Question

Find the sum of the series $(3^3-2^3)+(5^3-4^3)+(7^3-6^3)+\dots$ to

$\left(A\right)n$ terms

Answer 1

Genetal term of the series is in the form,

$T_k=(2k+1{)}^3-(2k{)}^3,\forall k\in N$

$\Rightarrow T_k=(2k+1-2k)\left[\right(2k+1{)}^2+(2k{)}^2+(2k+1\left)\right(2k\left)\right]$

$\Rightarrow T_k=1\times [4k^2+1+4k+4k^2+4k^2+2k]$

$\Rightarrow T_k=12k^3+6k+1$

$S_n=\sum _{k=1}^n{T}_k$

$S_n=\sum _{k=1}^n(12k^2+6k+1)$

$S_n=\left(12\sum _{k=1}^n{k}^2+6\sum _{k=1}^{n}k+\sum _{k=1}^n\right)$

$S_n=12\times \frac{n(n+1)(2n+1)}{6}+\frac{6\times n(n+1)}{2}+n$

$S_n=2n(n+1)(2n+1)+3n(n+1)+n$

$S_n=2n(2n^2+3n+1)+3n^3+3n+n$

$S_n=4n^3+6n^2+2n+3n^2+4n$

$\therefore S_n=4n^3+9n^2+6n$