Using the Binomial Theorem,evaluate (102)^{5} | vedclass

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Question

(A) We can express $102$ as $(100 + 2)$.Using the Binomial Theorem,$(a + b)^n = \sum_{k=0}^{n} {\binom{n}{k}} a^{n-k} b^k$.Here,$a = 100$,$b = 2$,and

Vedclass · Accepted Answer

$11040808032$

Vedclass · Answer

(A) We can express $102$ as $(100 + 2)$.Using the Binomial Theorem,$(a + b)^n = \sum_{k=0}^{n} {\binom{n}{k}} a^{n-k} b^k$.Here,$a = 100$,$b = 2$,and $n = 5$.$(100 + 2)^5 = \binom{5}{0}(100)^5 + \binom{5}{1}(100)^4(2) + \binom{5}{2}(100)^3(2)^2 + \binom{5}{3}(100)^2(2)^3 + \binom{5}{4}(100)(2)^4 + \binom{5}{5}(2)^5$.$= 1(10000000000) + 5(100000000)(2) + 10(1000000)(4) + 10(10000)(8) + 5(100)(16) + 1(32)$.$= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32$.$= 11040808032$.

Using the Binomial Theorem,evaluate $(102)^{5}$.

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