Question

Sum to n terms : 1.2.3.4+2.3.4.5+3.4.5.6+.....

Answer 1

Ah, nice.\[\sum_{k=1}^n k(k+1)(k+2)(k+3)\]\[= \frac{1}5 \sum_{k=1}^n (\color{red}{k+4} - (\color{blue}{k-1)})k(k+1)(k+2)(k+3) \]\[=\frac{1}5 \left(\sum_{k=1}^nk(k+1)(k+2)(k+3)(k+4)- \sum_{k=1}^n(k-1)k(k+1)(k+2)(k+3)\right) \]We can rewrite the second summation as\[ \sum_{k=1}^n(k-1)k(k+1)(k+2)(k+3) = \sum_{k=0}^{n-1} k(k+1)(k+2)(k+3)(k+4) \]So we're left with\[\frac{1}5 \left((a_1 + a_2+\cdots + a_n)- (a_0+a_1+\cdots + a_{n-1})\right)\]\[= \frac{1}5 (a_n - a_0)\]\[= \boxed{\frac{1}5n(n+1)(n+2)(n+3)(n+4) }\]