Question

The sum of series 1.2.4 + 2.3.5 + 3.4.6 + ----------- 20 terms.

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Answer 1

$n^{th}$ terms of the given series

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

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

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

Sum to n terms $s_n$ = ${\sum }_{i=1}^n$ $t_n$

$S_n$ = ${\sum }_{i=1}^n$ ($n^3$ + 4$n^2$ + 3n) = ${\sum }_{i=1}^n{n}^3$ + 4${\sum }_{i=1}^n{n}^2$ + 3${\sum }_{i=1}^{n}n$

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

$S_n$ =$\frac{n(n+1)}{12}$ [3n(n + 1) + 8(2n + 1) + 18]

$S_n$ =$\frac{n(n+1)}{12}$ [3$n^2$ + 19n + 26]

= $\frac{n(n+1)(n+2)(3n+13)}{12}$ --------------------(1)

Substituting n = 20, in the equation 1

$S_{20}$ = $\frac{20\left(21\right)\left(22\right)\left(73\right)}{12}$ = 56210