Fermat's Figurate Number Theory

Question

$$\sum_{i=1}^n \frac{i(i+1)...(i+k-1)}{k!}=\frac{n(n+1)...(n+k)}{(k+1)!}$$ Regard "k" as a fixed positive integer,

How can I prove this formula of Fermat by using induction on "n". I am not a mathematician. I am trying to understand the Fermat's approach to Quadratures, out of date form of Calculus.I have tried but not achieved.

Answer 1

There are two parts to a typical proof by mathematical induction.

1. the base case (here we may take $n=0$). That's easy in this case, both sides are $0$..

2. the induction step. Suppose it's true for $n$, and check that this implies it's true for $n+1$. Comparing the sum for $n$ and the sum for $n+1$, there is one additional term, the one for $i=n+1$. So here you want to check that $$ \frac{n(n+1)\ldots (n+k)}{(k+1)!} + \frac{(n+1)(n+2)\ldots(n+k)}{k!} = \frac{(n+1)(n+2)\ldots (n+k+1)}{(k+1)!} $$