If $n\in\mathbb N;$ show that $\frac{^nC_0}{x}-\frac{^nC_1}{x+1}+\frac{^nC_2}{x+2}-...+(-1)^n\frac{^nC_n}{x+n}=\frac{n!}{x(x+1)(x+2)...(x+n)}$

Question

If $n\in\mathbb N;$ show that $\frac{^nC_0}{x}-\frac{^nC_1}{x+1}+\frac{^nC_2}{x+2}-...+(-1)^n\frac{^nC_n}{x+n}=\frac{n!}{x(x+1)(x+2)...(x+n)}$

RHS$=\frac{n(n-1)!}{x(x+1)(x+2)...(x+n)}$

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

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

If I continue like this, I get the sense of LHS being formed.

What's the proper way to prove it?

Can we prove it starting with LHS?

Answer 1

Multiply $x(x+1)\cdots (x+n)$ to the both sides. Then you will find both sides being polynomials of degree $\leq n$. For polynomials of degree $\leq n$ to be identical, you need to check their values coincides for $n+1$ distinct values. We therefore substitute $x=-k, k= 0, 1, \dots, n$ to the both sides to see if they coincide.

When $x=-k$, we have

$$ LHS = (-1)^k \binom{n}{k}(-k)(-k+1)\cdots (-2)\cdot (-1)\cdot 1 \cdot 2 \cdots (n-k) = n! = RHS. $$

Since this is true for any $k=0, 1, 2, \dots, n$, we conclude that both sides are identical. Dividing both sides, by $x(x+1)\cdots(x+n)$, we obtain the desired formula.