Question:
Using Binomial Theorem, evaluate $(99)^{5}$
Solution:
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
$\therefore(99)^{5}=(100-1)^{5}$
$={ }^{5} \mathrm{C}_{0}(100)^{5}-{ }^{5} \mathrm{C}_{1}(100)^{4}(1)+{ }^{5} \mathrm{C}_{2}(100)^{3}(1)^{2}-{ }^{5} \mathrm{C}_{3}(100)^{2}(1)^{3}$
$+{ }^{5} \mathrm{C}_{4}(100)(1)^{4}-{ }^{5} \mathrm{C}_{5}(1)^{5}$
$=(100)^{5}-5(100)^{4}+10(100)^{3}-10(100)^{2}+5(100)-1$
$=10000000000-500000000+10000000-100000+500-1$
$=10010000500-500100001$
$=9509900499$