Using binomial theorem, prove that `(101)^(50) gt(100^(50)+99^(50)).`

Explanation

Putting `(101)^(50) = a and (100^(50)+99^(50)) = b,` we get
`(a-b)=(101)^(50)-(100)^(50)-(99)^(50)`
`(101)^(50) - (99)^(50)-(100)^(50)`
`(100+1)^(50)-(100-1)^(50)-(100)^(50)`
`2 xx[.^(50)C xx 100^(49) + .^(50)C_(3) xx 100^(47) +...+.^(50) C_(49) xx100]-(100)^(50)`
`=[2 xx.^(50) C_(3) xx 100^(47) + 2 xx.^(50)C_(5) xx 100^(45)+...+ 2 xx .^(50)C_(49)xx100]`
= (a positive integar).
Thus,`a-b gt0 rArr a gt b rArr (101)^(50) gt(100^(50) +99^(50)).`