Which is larger depends on what n is. It's clear that for n=1, 99^n + 100^n > 101^n
An easy way to justify that 101^n is larger than 99^n + 100^n once n=100, is to note 99^n + 100^n < 100^n + 100^n. Then, since 1.01^100 = 1+100*.01 + ... then 2<(101/100)^n for n<100, so 2*100^n < 101^n for n=100.
Basic D2 solution
