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
$$
\sum_{k=0}^{50}\binom{50}{k}(-1)^{50-k}100^k\quad\textrm{and}\quad \sum_{k=0}^{49}\binom{50}{k}100^k
$$
 A: 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 $k\gt1$:
$$
101^{50}-\left(100^{50}+99^{50}\right)=2\sum_{j=1}^{24}\binom{50}{2j+1}100^{49-2j}
$$
A: It could be amazing to compare
$$(2x)^x +(2 x-1)^x \qquad \text{and} \qquad (2 x+1)^x$$ or, better, their logarithms.
So, consider that you look for the zero of function
$$f(x)=\log \left((2x)^x +(2 x-1)^x\right)-\log \left((2 x+1)^x\right)$$ which has a trivial solution $x=2$.
So
$$x\gt 2 \qquad \implies\qquad (2x)^x +(2 x-1)^x \lt (2 x+1)^x$$
Checking
$$\left(
\begin{array}{cccc}
x &(2x)^x +(2 x-1)^x & (2 x+1)^x  &(2x)^x +(2 x-1)^x- (2 x+1)^x\\
 1 & 3 & 3 & 0 \\
 2 & 25 & 25 & 0 \\
 3 & 341 & 343 & -2 \\
 4 & 6497 & 6561 & -64 \\
 5 & 159049 & 161051 & -2002 \\
 6 & 4757545 & 4826809 & -69264 \\
\end{array}
\right)$$
May be, you could use induction.
A: $$\sum_{k=0}^{49}\binom{50}{k}100^k$$$$=50\cdot100^{49}+1225\cdot100^{48}+  ...$$$$>62\cdot100^{49}$$$$>99^{50}$$
The last inequality comes from $$\ln{62} + 49\ln{100}=229.78...>50\ln{99}=229.75...$$
