Summation of series by method of differences

Question

1.5.9+2.6.10+3.7.11 ... n terms

N th term would be n(n+4)(n+8)

My try

N th term =n[(n+1)+3)][n+2+(6)]

I am unable to break the series further .

How can I simplify this further using method of differences ?

Answer 1

To use the method of differences, we have to find $x_n=an^4+bn^3+cn^2+dn$ such that $x_{n+1}-x_n=n(n+4)(n+8)$. Then we will have

$$\sum_{k=1}^nk(k+4)(k+8)=\sum_{k=1}^n(x_{k+1}-x_k)=x_{n+1}-x_1$$

What we have to do is to determine $a$, $b$, $c$ and $d$. By comparing the coefficients, we have

$$x_n=\frac{1}{4}n^4+\frac{7}{2}n^3+\frac{41}{4}n^2-14n$$

Therefore,

\begin{align} \sum_{k=1}^nk(k+4)(k+8)&=\frac{1}{4}[(n+1)^4-1]+\frac{7}{2}[(n+1)^3-1]\\ &\qquad\qquad+\frac{41}{4}[(n+1)^2-1]-14[(n+1)-1]\\ &=\frac{n}{4}[(n^3+4n^2+6n+4)+14(n^2+3n+3)+41(n+2)-56]\\ &=\frac{n}{4}(n^3 + 18 n^2 + 89 n + 72)\\ &=\frac{1}{4}n(n + 1) (n + 8) (n + 9) \end{align}

Answer 2

$$\sum_{k=1}^nk(k+4)(k+8)=\sum_{k=1}^n(k^3+12k^2+32k)=$$ $$=\frac{n^2(n+1)^2}{4}+12\cdot\frac{n(n+1)(2n+1)}{6}+32\cdot\frac{n(n+1)}{2}=$$ $$=\frac{n(n+1)}{4}(n^2+n+8(2n+1)+64)=\frac{n(n+1)(n^2+17n+72)}{4}.$$