Binomial summations 
The given answer is zero. I tried adjusting the 
coefficients, changing to binomial co-efficients.
Like $99C_r$but still the inside terms, i am not able 
to adjust.
 A: 
We  obtain
  \begin{align*}
\color{blue}{99^{50}}&\color{blue}{-99\cdot98^{50}+\frac{99\cdot98}{1\cdot 2}\cdot97^{50}
-\cdots+99}\\
&=\sum_{j=0}^{98}\binom{99}{j}(-1)^j(99-j)^{50}\tag{1}\\
&=-\sum_{j=0}^{99}\binom{99}{j}(-1)^jj^{50}\tag{2}\\
&=-\sum_{j=0}^{99}\binom{99}{j}(-1)^j50![z^{50}]e^{jz}\tag{3}\\
&=-50![z^{50}]\sum_{j=0}^{99}\binom{99}{j}\left(-e^z\right)^j\tag{4}\\
&=-50![z^{50}](1-e^z)^{99}\tag{5}\\
&=-50![z^{50}]\left(1-\left(1+z+\frac{z^2}{2}+\cdots\right)\right)^{99}\tag{6}\\
&=50![z^{50}]\left(z+\frac{z^2}{2}+\cdots\right)^{99}\tag{7}\\
&\,\,\color{blue}{=0}
\end{align*}

Comment:


*

*In (1) we write the expression using sigma notation and binomial coefficients.

*In (2) we add $0$ by setting the upper limit to $99$ and change the order of the sum $j\to 99-j$.

*In (3) we use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series and we note that
$$j^n=n![z^n]e^{jz}=n![z^n]\sum_{k=0}^\infty\frac{(jz)^{k}}{k!}$$

*In (4) we do some rearrangements as preparation for the next step.

*In (5) we apply the binomial theorem.

*In (6) we expand the exponential series to better see what's going on.

*In (7) we simplify the expression and observe that the series starts with powers in $z\geq 99$, so that the coefficient of $z^{50}$ is zero.
