Question

Find the sum of $n$ terms of the series whose $n^{th}$ term is $(n^2+5n+4)(n^2+5n+8)$.

Answer 1

$T_n=(n^2+5n+4)(n^2+5n+8)=(n+1)(n+4)\left\{\right(n+2\left)\right(n+3)+2\}$

$=(n+1)(n+2)(n+3)(n+4)+2(n^2+5n+4)$

$T_n=\frac{1}{5}(n+1)(n+2)(n+3)(n+4)\left\{\right(n+5)-n\}+2(n^2+5n+4)$

$A_n=(n+1)(n+2)(n+3)(n+4)=\frac{1}{5}(n+1)(n+2)(n+3)(n+4)\left\{\right(n+4)-n\}$

$A_n=\frac{1}{5}\left\{\right(n+1\left)\right(n+2\left)\right(n+3\left)\right(n+4\left)\right(n+5)-n(n+1\left)\right(n+2\left)\right(n+3\left)\right(n+4\left)\right\}$

$A_1=\frac{1}{5}[2\cdot 3\cdot 4\cdot 5\cdot 6-1\cdot 2\cdot 3\cdot 4\cdot 5]$

$A_2=\frac{1}{5}[3\cdot 4\cdot 5\cdot 6\cdot 7-2\cdot 3\cdot 4\cdot 5\cdot 6]$

$A_3=\frac{1}{5}[4\cdot 5\cdot 6\cdot 7\cdot 8-3\cdot 4\cdot 5\cdot 6\cdot 7]$

$\dots \dots \dots \dots \dots \dots$

$\dots \dots \dots \dots \dots \dots$

$A_3=\frac{1}{5}\left\{\right(n+1\left)\right(n+2\left)\right(n+3\left)\right(n+4\left)\right(n+5)-n(n+1\left)\right(n+2\left)\right(n+3\left)\right(n+4\left)\right\}$

$S_1=A_1+A_2+A_3+\cdots +A_n=\frac{1}{5}(n+1)(n+2)(n+3)(n+4)(n+5)-24$

$S_n=S_1+2\sum _{n=1}^n{n}^2+10\sum _{n=1}^{n}n+8n$

$S_n=\frac{1}{5}(n+1)(n+2)(n+3)(n+4)(n+5)-24+\frac{1}{3}n(n+1)(2n+1)+5n(n+1)+8n$

$S_n=\frac{1}{15}(n+1)(n+2)(3n^3+36n^2+151n+240)-32$