&#10;&#13;&#10;\Rightarrow \displaystyle\sum_{r=1}^{k+1} r(r+1)(r+2)(r+3)(r+4) = \frac{1}{6}k(k+1)(k+2)(k+3)(k+4)  (k+5) + (k+1)(k+2)(k+3)(k+4)(k+5)