Question

A = (99^50 + 100^50) and B = 101^50, the larger is?

a)" "A

b)" "B

c)Both are same

d)Cant Say

Solution

We can solve this problem using binomial concepts.

A = (99^50 + 100^50) = (100-1)^50 + 100^50

and B = 101^50 = (100+1)^50

B-A = (100+1)^50 - ((100-1)^50 + 100^50)

B-A = 100^50 + 50C_1(100)^49 + 50C_2(100)^48 .......... - (100^50 - 50C_1(100)^49 + 50C_2(100)^48 - 50C_3(100)^47 .......... + 100^50)

B-A = 50C_1(100)^49 + 50C_2(100)^48 .......... - (100^50 - 50C_1(100)^49 + 50C_2(100)^48 - 50C_3(100)^47 ..........)

B-A = 50C_1(100)^49 + 50C_2(100)^48 .......... - 100^50 + 50C_1(100)^49 - 50C_2(100)^48 + 50C_3(100)^47 ..........

B-A = 2xx50C_1(100)^49 + 50C_2(100)^48 .......... - 100^50 - 50C_2(100)^48 + 50C_3(100)^47 ..........

B-A = 100xx(100)^49 + 50C_2(100)^48 .......... - 100^50 - 50C_2(100)^48 + 50C_3(100)^47 ..........

B-A = (100)^50 + 50C_2(100)^48 .......... - 100^50 - 50C_2(100)^48 + 50C_3(100)^47 ..........

B-A = 50C_2(100)^48 +50C_3(100)^47 .......... - 50C_2(100)^48 + 50C_3(100)^47 ..........

B-A = (50C_3(100)^47 + 50C_4(100)^46 ..........) + 50C_3(100)^47 - 50C_4(100)^46 ..........

Ignoring higher order terms we can see that B-A > 1, hence B is bigger.