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Q. The value of $(98)^5$ is

Binomial Theorem

Solution:

We express 98 as the sum or difference of two numbers whose powers are easier to calculate and then use binomial theorem.
Write $98=100-2$
Therefore, $(98)^5=(100-2)^5$
$={ }^5 C_0(100)^5-{ }^5 C_1(100)^4 \cdot 2+{ }^5 C_2(100)^3 2^2$
$ -{ }^5 C_3(100)^2(2)^3+{ }^5 C_4(100)(2)^4-{ }^5 C_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 .$