Can someone please point out which induction's step I did wrong?

Question

$$\sum_{k=1}^n k(k+1)(k+2)...(k+p-1)= \frac{n(n+1)(n+2)...(n+p)} {p+1}$$

induction by substitute n+1: $$(n+p+1)!= (n+p)!+(p+1)(n+p)! $$ I couldnt figure out how to countinue after this step. Thanks

Answer 1

Induction step: $$\sum_{k=1}^{n+1} k(k+1)(k+2)...(k+p-1)=\\ \sum_{k=1}^n k(k+1)(k+2)...(k+p-1)+(n+1)(n+2)\cdots(n+p)=\\ \frac{n(n+1)(n+2)...(n+p)} {p+1}+(n+1)(n+2)\cdots(n+p)=\\ \frac{(n+1)(n+1+1)...(n+1+p-1)(n+1+p)} {p+1}.$$

Answer 2

The expression $$(n+p+1)!= (n+p)!+(p+1)(n+p)!$$

should have been $$(n+p+1)!= (n+p+1)(n+p)! =n(n+p)!+(p+1)(n+p)!$$

Answer 3

Suppose you have the answer for $n$. For $n+1$, the sum will be $\sum\limits_{k=1}^{n+1} k(k+1)...(k+p-1)$
That is $\sum\limits_{k=1}^{n} k(k+1)...(k+p-1) \ + (n+1)(n+2)...(n+p)$
Then, by induction hypothesis, you have $\frac{n(n+1)...(n+p)}{p+1}+(n+1)(n+2)...(n+p)=\frac{n(n+1)...(n+p) + (n+1)(n+2)...(n+p)(p+1)}{p+1}=\frac{((n+1)(n+2)...(n+p))(n + p + 1)}{p+1}$
and you have your answer. Of course, you should also write down the basis for the induction.