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Q. Compute $(98)^5$.

Binomial Theorem

Solution:

We have $(98)^5 = (100 - 2)^5$
$= \,{}^5C_0(100)^5- \,{}^5C_1(100)^4 \cdot 2+\,{}^5C_2(100)^32^2- \,{}^5C_3(100)^2(2)^3$
$+\,{}^5C_4(100)(2)^4 - \,{}^5C_5(2)^5$
$= 10000000000-(5 \times 100000000 \times 2)+(10 \times 1000000 \times 4)$
$- (10 \times 10000 \times 8) + (5 \times 100 \times 16) - 32$
$= 10040008000 - 1000800032$
$ = 9039207968$.