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

279 155 Binomial Theorem Report Error

A

903920986

3%

B

9039207968

93%

C

309820968

2%

D

None of the above

2%

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, $(98)^{5} = (100 - 2)^{5}$
$=^{5} C_{0} (100)^{5} -^{5} C_{1} (100)^{4} \cdot 2 +^{5} C_{2} (100)^{3} 2^{2}$
$-^{5} C_{3} (100)^{2} (2)^{3} +^{5} C_{4} (100) (2)^{4} -^{5} C_{5} (2)^{5}$
$= 10000000000 - 5 \times 100000000 \times 2 + 10 \times 1000000 \times 4$
$- 10 \times 10000 \times 8 + 5 \times 100 \times 16 - 32$
$= 10040008000 - 1000800032 = 9039207968.$