Binomial summations

Question

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.

Answer 1

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^{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.