Using binomial theorem, prove that `(101)^(50)> 100^(50)+99^(50)dot`

Explanation

Since , ` (101)^(50) - (99)^(50) = (100 + 1)^(50) - (100 -1)^(50)`
` 2 {""^(50)C_(1) (100)^(49) + ""^(50)C_(3) (100)^(47) + ""^(50)C_(5) (1000)^(45) + …}`
` = 2 xx ""^(50)C_(1) (100)^(49) + 2 {""^(50)C_(3) (100)^(47) + ""^(50)C_(5) (100)^(45) + ...}`
` = (100)^(50) + "(a positive number)" gt (100)^(50)`
Hence , ` (101)^(50) - (99)^(50) gt (100)^(50)`
` rArr (101)^(50) gt (100)^(50) + (99)^(50)` .