Question

Find the sum to n terms

$1\times 2\times 3+2\times 3\times 4+3\times 4\times 5+\dots \dots$

Answer 1

The given series in $1\times 2\times 3+2\times 3\times 4+3\times 4\times 5+\dots$ to n terms. Let $a_n$ be the nth term of the given series and $S_n$ be the sum of 'n' terms of given series.

$\therefore a_n=$ [nth term of 1, 2, 3, ....] [nth term of 2, 3, 4, ....] [nth term of 3, 4, 5, ...]

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

$\therefore S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n(k^3+3k^2+2k)$

$=[1^3+{3.1}^2+2.1]+[2^3+{3.2}^2+2.2]+\dots \dots +[n^3+3n^2+2n]$

$=[1^3+2^3+\dots \dots +n^3]+3[1^2+2^2+\dots \dots +n^2]+2[1+2+\dots \dots +n]$

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

$=\frac{n^2(n+1{)}^2}{4}+\frac{n(n+1)(2n+1)}{2}+n(n+1)$

$=n(n+1)[\frac{n(n+1)}{4}+\frac{(2n+1)}{2}+1]$

$=n(n+1)\left[\frac{n^2+n+4n+2+4}{4}\right]$ $=\frac{n(n+1)(n^2+5n+6)}{4}$ $=\frac{1}{4}n(n+1)(n+2)(n+3)$.