Find the value of $ 99^{50}-\binom{99}{1}(98)^{50}+\binom{99}{2}(97)^{50}-\cdots \cdots +99$ Find the value of $\displaystyle 99^{50}-\binom{99}{1}(98)^{50}+\binom{99}{2}(97)^{50}-\cdots \cdots +99$
Binomial identity:
$\displaystyle (1-x)^{99} = \binom{99}{0}-\binom{99}{1}x+\binom{99}{2}x^2-\cdots \cdots -\binom{99}{99}x^{99}$
I want be able to go further, could some help me with this, thanks

It is simply zero. Let $\delta$ be the (difference) operator mapping $p(x)$ into $p(x)-p(x-1)$: if $p(x)$ is a polynomial with degree $d\geq 1$, then $(\delta p)(x)$ is a polinomial with degree $d-1$. Since
$$(\delta^n p)(x) = \sum_{k=0}^{n}(-1)^k \binom{n}{k} p(x-k) \tag{1}$$
if $p(x)$ is a polynomial with degree $<n$, the RHS of $(1)$ is constantly zero. 
The given sum is the RHS of $(1)$ in the case $p(x)=x^{50}$, $n=99=x$.
