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

• the $-$ and $+$ are alternating or not? Commented Jan 10, 2017 at 11:36
• yes @Exodd - and + are alternate.
– DXT
Commented Jan 10, 2017 at 11:37
• If your signs are alternating, this looks like an inclusion-exclusion thing -- but it's a bit strange because it seems to try to compute "the number of length-50 sequences drawn from 99 symbols that don't miss any of the 99 symbols", which is impossible. Commented Jan 10, 2017 at 11:42
• @Rohan -- to fit the pattern it should be $\binom{99}{98}1^{50}=99$. Commented Jan 10, 2017 at 11:45
• @Rohan: It's easier to write the general term as $(-1)^{n+1}\binom{99}{n} n^{50}$, with $n$ decreasing from $99$ to $1$. Then the last term is $\binom{99}{1} 1^{50} = 99$, as written.. Commented Jan 10, 2017 at 11:47

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 $p(x)=x^{50}$, $n=99=x$.