Question

Find the sum of series to $n$ terms:
$1\cdot 2\cdot 5+2\cdot 3\cdot 6+3\cdot 4\cdot 7+.....$

Answer 1

$1\cdot 2\cdot 5+2\cdot 3\cdot 6+3\cdot 4\cdot 7+.....$

In each term of sequence, first, second and third number forms an A.P. with common difference $1,1$ and $1$ respectively.

Let $T_n$ be the $n^{th}$ term of the given series, then
$T_n=\{n^{th}\text{term of}1,2,3,...\}\times \{n^{th}\text{term of}2,3,4,...\}\times \{n^{th}\text{term of}5,6,7,...\}$

$=\{1+(n-1)\times 1\}\times \{2+(n-1)\times 1\}\times \{5+(n-1)\times 1\}$

$=n(n+1)(n+4)$

$=(n^2+n)(n+4)$

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

$T_n=n^3+5n^2+4n\cdots \left(1\right)$

Now, let $S_n$ ne the sum of $n$ terms of the given series

We have:
$S_n=\sum T_n$

$=\sum (n^3+5n^2+4n)\left[\text{Using equation}\right(1\left)\right]$

$=\sum n^3+5\sum n^2+4\sum n$

$=\frac{n^2{(n+1)}^2}{4}+\frac{5n(n+1)(2n+1)}{6}+\frac{4n(n+1)}{2}$

$=n(n+1)\{\frac{n(n+1)}{4}+\frac{5}{6}(2n+1)+2\}$

$=n(n+1)\left\{\frac{3n(n+1)+2\times 5(2n+1)+2\times 12}{12}\right\}$

$=n(n+1)\left\{\frac{3n^2+3n+20n+10+24}{12}\right\}$

$=\frac{n}{12}(n+1)(3n^2+23n+34)$

$\therefore \text{Sum of series}{S}_n=\frac{n}{12}(n+1)(3n^2+23n+34)$