find sum of n terms of the sequence 1.3.(2^2) + 2.4.(3^2)+3.5.(4^2)+.......

Question

S=$1.3.2^2+2.4.3^2+3.5.4^2+........$ upto n terms

Find S?

This was asked under the heading using method of difference and ans given was (n)(n+1)(n+2)(n+3)(2n+3)/10

Answer 1

The brute force way: $$\sum_{k=1}^nk(k+1)^2(k+2)=\sum_{k=1}^n(k^4+4k^3+5k^2+2k)=$$ $$=\tfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}+4\cdot\frac{n^2(n+1)^2}{4}+5\cdot\frac{n(n+1)(2n+1)}{6}+2\cdot\frac{n(n+1)}{2}=$$ $$=\frac{n(n+1)(n+2)(n+3)(2n+3)}{10}.$$ Now, easy to make the telescopic sum. $$\sum_{k=1}^nk(k+1)^2(k+2)=$$ $$=\sum_{k=1}^n\left(\frac{k(k+1)(k+2)(k+3)(2k+3)}{10}-\frac{(k-1)k(k+1)(k+2)(2k+1)}{10}\right)=$$ $$=\frac{n(n+1)(n+2)(n+3)(2n+3)}{10}.$$

Answer 2

The $0^{th}$, $-1^{st}$ and $-2^{nd}$ terms are all zero, so think of the sum as zero after $-3,-2,-1$ and $0$ terms. So the sum, which is a polynomial, is a multiple of $n(n+1)(n+2)(n+3)$