Question

I really stuck on trying to prove this

n
E r(r+1)(r+2)...(r+p-1)=(1/p+1)n(n+1)(n+2)...(n+p)
r=1

Answer 1

`\displaystyle\sum_(r=1)^nr(r+1)\cdots(r+p-1)`

When
`n=1`,


`\displaystyle\sum_(r=1)^1r(r+1)\cdots(r+p-1)=1(1+1)(1+2)\cdots(1+p-2)(1+p-1)=p!`

Meanwhile, you have on the right


`((1)(1+1)(1+2)\cdots(1+p-2)(1+p-1)(1+p))/(p+1)=(1)(1+1)(1+2)\cdots(p-1)(p)=p!`

so the equality holds for
`n=1`.

Assume it holds for
`n=k`, i.e. that


`\displaystyle\sum_(r=1)^kr(r+1)\cdots(r+p-1)=(k(k+1)(k+2)\cdots(k+p-1)(k+p))/(p+1)`

Now for
`n=k+1`, you have


`\displaystyle\sum_(r=1)^(k+1)r(r+1)\cdots(r+p-1)=\sum_(r=1)^kr(r+1)\cdots(r+p-1)+(k+1)(k+2)\cdots(k+1+p-1)`

`=\displaystyle(k(k+1)(k+2)\cdots(k+p-1)(k+p))/(p+1)+(k+1)(k+2)\cdots(k+1+p-2)(k+1+p-1)`

`=\displaystyle(k(k+1)(k+2)\cdots(k+p-1)(k+p))/(p+1)+(k+1)(k+2)\cdots(k+p-1)(k+p)`

`=\left(\frac k{p+1}+1\right)(k+1)(k+2)\cdots(k+p-1)(k+p)`

`=(k+p+1)/(p+1)(k+1)(k+2)\cdots(k+p-1)(k+p)`

`=((k+1)(k+2)\cdots(k+p-1)(k+p)(k+p+1))/(p+1)`

as required.