Which number is larger? Using Binomial Theorem

Question

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 $$ \sum_{k=0}^{50}\binom{50}{k}(-1)^{50-k}100^k\quad\textrm{and}\quad \sum_{k=0}^{49}\binom{50}{k}100^k $$

Answer 1

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 $k\gt1$: $$ 101^{50}-\left(100^{50}+99^{50}\right)=2\sum_{j=1}^{24}\binom{50}{2j+1}100^{49-2j} $$

Answer 2

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.

Answer 3

$$\sum_{k=0}^{49}\binom{50}{k}100^k$$$$=50\cdot100^{49}+1225\cdot100^{48}+ ...$$$$>62\cdot100^{49}$$$$>99^{50}$$

The last inequality comes from $$\ln{62} + 49\ln{100}=229.78...>50\ln{99}=229.75...$$