Now,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 Thus, (98)^5=(100-2)^5 = ^5 C_0(100)^5- ^5 C_1(100)^4 .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 =100000000000-5 100000000 2+10 1000000 4-10 10000 8+5 100 16-32 =10040008000-1000800032=9039207968.

Compute (98)^5.

Question

Compute $(98)^5$.

✓

Answer

Now,we express 98 as the sum or difference of two numbers whose powers are easier to calculate, and then use Binomial Theorem.
$\text { Write } 98=100-2$
$\text { Thus, }(98)^5=(100-2)^5$
$={ }^5 \mathrm{C}_0(100)^5-{ }^5 \mathrm{C}_1(100)^4 .2+{ }^5 \mathrm{C}_2(100)^3 2^2-{ }^5 \mathrm{C}_3(100)^2(2)^3+{ }^5 \mathrm{C}_4(100)(2)^4-{ }^5 \mathrm{C}_5(2)^5$
$=100000000000-5 \times 100000000 \times 2+10 \times 1000000 \times 4-10 \times 10000 \times 8+5 \times 100 \times 16-32$
$=10040008000-1000800032=9039207968$.

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