By using binomial theorem, expand the following:
$(99)^{5}$

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Solution

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

Using binomial theorem, we have
$(99)^{5} = 5_{C_{0}} (100)^{5} + 5_{C_{1}} (100)^{4} (- 1) + 5_{C_{2}} (100)^{3} (- 1)^{2} + 5_{C_{3}} (100)^{2} (- 1)^{3}$

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

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

$= 10010000500 - 500100001$

$= 9509900499$

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