Help on induction, couldn't make both side the same value

Question

$$\sum_{k=1}^n k(k+1)(k+2)...(k+p-1)= \frac{n(n+1)(n+2)...(n+p)} {p+1} $$ Induction g(n+1)=f(n+1)+g(n) sub n+1
(1)$$(n+1)(n+2)...(n+p)(p+1)+\frac{n(n+1)(n+2)...(n+p)} {p+1}(p+1) $$ (2) $$\frac{(n+1)(n+2)...(n+1+p)} {p+1}(p+1)$$ make (1)=(2) $$(n+p)!(p+1)+(n+p)!=(n+1+p)! $$

I then couldnt make the R.H.S==L.H.S

Answer 1

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

First of all this given sum can be simplified by using factorials

$$\frac{(n+p)!}{(n-1)!}~=~n(n+1)(n+2)\cdots(n+p)$$

In general the term

$$(x,n)~=~x^{\overline{n}}~=~\frac{(x+n+1)!}{(x+1)!}$$ is called Pochhammer symbol or rising factorial. $$\begin{align}\sum_{k=1}^n \frac{(k+p-1)!}{(k-1)!}~&=~\frac{n(n+1)(n+2)...(n+p)} {p+1}\\~&=~\frac{(n+p)!}{(n-1)!(p+1)}\end{align}$$ $$\begin{align}\sum_{k=1}^{n+1} \frac{(k+p-1)!}{(k-1)!}~&=~\frac{(n+1)(n+2)(n+3)...(n+p+1)} {p+1}\\~&=~\frac{(n+p+1)!}{n!(p+1)}\end{align}$$ $$\begin{align}\sum_{k=1}^{n+1} \frac{(k+p-1)!}{(k-1)!}~=~&\sum_{k=1}^n \frac{(k+p-1)!}{(k-1)!}~+~\frac{(n+p)}{n!}\end{align}$$ $$\begin{align}\sum_{k=1}^n \frac{(k+p-1)!}{(k-1)!}~+~\frac{(n+p)!}{n!}~&=~\frac{(n+p)!}{(n-1)!(p+1)}~+~\frac{(n+p)!}{n!}\\&=~\frac{n(n+p)!+(p+1)(n+p)!}{n!(p+1)}\\&=~\frac{(n+p+1)(n+p)!}{n!(p+1)}\\&=~\frac{(n+p+1)!}{n!(p+1)}\end{align}$$

Maybe this slightly different approach to your problem can help you to get further.

Answer 2

I'm not sure how or why you are writing 1).

Just do the following.

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

Then $\sum_{k=1}^{n+1}k(k+1)...(k+p-1) = [\sum_{k=1}^n k(k+1)...(k+p-1)]+ (n+1)((n+1)+1)...((n+1)+p-1)$

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

$= [(n+1)(n+2).....(n+p)](\frac n{p+1} + 1)$

$=[(n+1)(n+2).....(n+p)](\frac n{p+1} + \frac{p+1}{p+1})$

$=[(n+1)(n+2).....(n+p)](\frac {n+p + 1}{p+1})$

$=\frac {(n+1).....(n+1 + p)}{p+1}$