Using Binomial Theorem, evaluate (99)5

99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.

It can be written that, 99 = 100 – 1

$\therefore(99)^{5}=(100-1)^{5}$

$={ }^{5} \mathrm{C}_{0}(100)^{5}-{ }^{5} \mathrm{C}_{1}(100)^{4}(1)+{ }^{5} \mathrm{C}_{2}(100)^{3}(1)^{2}-{ }^{5} \mathrm{C}_{3}(100)^{2}(1)^{3}$

$+{ }^{5} \mathrm{C}_{4}(100)(1)^{4}-{ }^{5} \mathrm{C}_{5}(1)^{5}$

$=(100)^{5}-5(100)^{4}+10(100)^{3}-10(100)^{2}+5(100)-1$

$=10000000000-500000000+10000000-100000+500-1$

$=10010000500-500100001$

$=9509900499$