Question

Find the sum of the following series to n terms:
$\mathrm{1.2.4}+\mathrm{2.3.7}+\mathrm{3.4.10}+...$

Answer 1

Let $T_n$ be the terms of the given series.

Then

$T_n(n^{th}termof1,2,\mathrm{3...})\times (n^{th}termof2,3,\mathrm{4...})\times (n^{th}termof2,3,\mathrm{4...})$

$=[1+(n-1)\times 1].[2+(n-1)\times 1][4+(n-1)\times 3]$

$=[1+n-1][2+n-1][4+3n-3]$

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

$=3n^3+n^2+3n^2+n$

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

$T_n=3n^3+4n^2+n$

Let $S_n$ denoted sum of $n$ terms of the given series

$S_n=\sum _{n=1}^n{T}_n=\sum _{n=1}^n{n}^2+\sum _{n=1}^{n}n$

$=\frac{3}{1}{\left[\frac{n(n+1)}{2}\right]}^2+4\left[\frac{n(n+1)(2n+1)}{6}\right]+\frac{n(n+1)}{2}$

$=\frac{3}{4}\left[n\right(n+1){]}^2+\frac{2n(n+1)(2n+1)}{3}+\frac{n(n+1)}{2}$

$=\frac{n}{12}(n+1)[9n^2+9n+16n+8+6]$

$S_n=\frac{n}{12}(n+1)[9n^2+25n+14]$