Which is larger : `(99^(50)+100^(50))` or `(101)^(50)`.

Explanation

We have
`101^(50) = (100 + 1)^(50) = .^(50)C_(0)100^(50) + .^(50)C_(1) 100^(49) + .^(50)C_(2)100^(48) + "…"`
` = 100^(50) + 50 xx 100^(49) + (50 xx 49)/(1 xx 2) x 100^(48)`
` + ( 50 xx 49 xx 48)/(1 xx 2 xx 3) xx 100^(47)"….."(1)`
and `99^(50) = (100 - 1)^(50)`
`= .^(50)C_(0)100^(50) - .^(50)C_(1)100^(49) + .^(50)C_(2)100^(48)+"...."`
` = 100^(50) - 50 xx 100^(49) + (50 xx 49)/(1 xx 2) xx 100^(48)`
`- (50 xx 49 xx 48)/(1 xx 2 xx 3) xx 100^(47)"......"(2)`
Substracting (2) form (1), we get
`101^(50) - 99^(50) = 100^(50) + 2 xx (50 xx 49 xx 48)/(1 xx 2 xx 3) xx 100^(47) + "......." gt 100^(50)`
Hence, `101^(50) gt 100^(50) + 99^(50)`.