Using the Binomial Theorem,evaluate (99)^{5}. | vedclass

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(A) We can express $99$ as $(100 - 1)$.Using the Binomial Theorem,$(a - b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^{k}$.$(99)^{5} = (100 - 1)^{

Vedclass · Accepted Answer

$9509900499$

Vedclass · Answer

(A) We can express $99$ as $(100 - 1)$.Using the Binomial Theorem,$(a - b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^{k}$.$(99)^{5} = (100 - 1)^{5} = \binom{5}{0}(100)^{5} - \binom{5}{1}(100)^{4} + \binom{5}{2}(100)^{3} - \binom{5}{3}(100)^{2} + \binom{5}{4}(100)^{1} - \binom{5}{5}(1)^{5}$.$= 1(10000000000) - 5(100000000) + 10(1000000) - 10(10000) + 5(100) - 1$.$= 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1$.$= 9500000000 + 9900000 + 499$.$= 9509900499$.

Using the Binomial Theorem,evaluate $(99)^{5}$.

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