Which number is larger? Using Binomial Theorem Which expression is larger,
$$
99^{50}+100^{50}\quad\textrm{ or }\quad 101^{50}?
$$
Idea is to use the Binomial Theorem:
The right hand side then becomes
$$
101^{50}=(100+1)^{50}=\sum_{k=0}^{50}\binom{50}{k}1^{50-k}100^k=100^{50}+\sum_{k=0}^{49}\binom{50}{k}100^k
$$
The left hand side reads
$$
99^{50}+100^{50}=(100-1)^{50}+100^{50}=\sum_{k=0}^{50}\binom{50}{k}(-1)^{50-k}100^k+100^{50}
$$
Thus, since both sides have the summand $100^{50}$, it remains to compare
$$
\sum_{k=0}^{50}\binom{50}{k}(-1)^{50-k}100^k\quad\textrm{and}\quad \sum_{k=0}^{49}\binom{50}{k}100^k
$$

The binomial theorem says
$$
101^{50}=\sum_{k=0}^{50}(+1)^k\binom{50}{k}100^{50-k}
$$
and
$$
99^{50}=\sum_{k=0}^{50}(-1)^k\binom{50}{k}100^{50-k}
$$
Furthermore, $100^{50}=2\binom{50}{1}100^{49}$, which is the difference in the $k=1$ terms. Thus, the difference is the sum of the differences of the odd terms for $k\gt1$:
$$
101^{50}-\left(100^{50}+99^{50}\right)=2\sum_{j=1}^{24}\binom{50}{2j+1}100^{49-2j}
$$
