Question

Find the sum of n terms of the series whose general term is $n^4+6n^3+5n^2$.

Answer 1

Assuming $n^4+6n^3+5n^2=A+Bn+Cn(n+1)+Dn(n+1)(n+2)+En(n+1)(n+2)(n+3)$

Equating the coefficient of the like powers on $n$, we get

$A=0,B=0,C=-6,D=0,E=1$

$\therefore T_n=n^4+6n^3+5n^2=n(n+1)(n+2)(n+3)-6n(n+1)=\frac{1}{5}n(n+1)(n+2)(n+3)\left\{\right(n+4)-(n-1\left)\right\}-2n(n+1)\left\{\right(n+2)-(n-1\left)\right\}$

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

$T_1=\frac{1}{5}[1\cdot 2\cdot 3\cdot 4\cdot 5-0]-2[1\cdot 2\cdot 3-0]$

$T_2=\frac{1}{5}[2\cdot 3\cdot 4\cdot 5\cdot 6-1\cdot 2\cdot 3\cdot 4\cdot 5]-2[2\cdot 3\cdot 4-1\cdot 2\cdot 3]$

$T_3=\frac{1}{5}[3\cdot 4\cdot 5\cdot 6\cdot 7-2\cdot 3\cdot 4\cdot 5\cdot 6]-2[3\cdot 4\cdot 5-2\cdot 3\cdot 4]$

$\dots \dots \dots \dots$

$\dots \dots \dots \dots$

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

$S_n=T_1+T_2+T_3+\dots \dots +T_n=\frac{1}{5}n(n+1)(n+2)(n+3)(n+4)-2n(n+1)(n+2)$

$S_n=\frac{1}{5}n(n+1)(n+2)(n^2+7n+2)$