Question

Find the $n^{th}$ term and the sum of n terms of the series $30,144,420,960,1890,3360,....$

Answer 1

$\Rightarrow 301444209601890\mathrm{3360....}$

$114276540930\mathrm{1470....}$

$162264390\mathrm{540....}$

$102126\mathrm{150....}$

$24\mathrm{24....}$

$\therefore u_n=30+114(n-1)+81(n-1)(n-2)+17(n-1)(n-2)(n-3)+(n-1)(n-2)(n-3)(n-4);.....$

$=n(n+1)(n^2+6n+8)$

$n(n+1)(n+2)(n+4)$

$\therefore S_n=\frac{1}{20}n(n+1)(n+2)(n+3)(4n+21)$

Sum can be obtained by;-

$u_n=n(n+1)(n+2)(n+4)$

$S_n=\frac{1}{5.1}n(n+1)(n+2)(n+3)(n+4).+C.$

$C=30-24=6$

$S_n=\frac{1}{5}n(n+1)(n+2)(n+3)(n+4)+6$

$S_n=\frac{1}{5}n(n+1)(n+2)(n+3)(4n+21)$