An algebra problem by A. R Apon

$\begin{aligned} 101^{50} & = (100+1)^{50} = 100^{50} + 50\times 100^{49} + \frac {50\times 49}2 \times 100^{48} + \cdots + 1 \\ 99^{50} & = (100-1)^{50} = 100^{50} - 50\times 100^{49} + \frac {50\times 49}2 \times 100^{48} - \cdots + 1 \\ 101^{50} - 99^{50} & = 100^{50} + \frac {50\times 49\times 48}3 \times 100^{47} + \frac {50\times 49\times 48\times 47 \times 46 }{60} \times 100^{45} + \cdots + 100^2 \\ \implies 101^{50} - 99^{50} & > 100^{50} \\ 101^{50} & > 99^{50} + 100^{50} \end{aligned}$

101^50 is greater than (99^50 + 100^50)

As 101^50= 1.645 10^100 and (99^50+100^50)= 1.605 10^100

Thank You