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

Answer

N/a

Explanation

`(105)^(50)-(100^(50)+90^(50))`
`=101^(50)-100^(50)-99^(50)`
`=(100+1)^(50)-(100-1)^(50)-100^(50)`
`=[100^(50+50)C_(1).100^(49)+^(50)C_(2).100^(40)`
`.......+^(50)C_(49).100+1]-[100^(50)`
`-^(50)C_(1).100^(49)+^(50)C_(2).100^(48)`
`+....-^(50)C_(49).100+1]-100^(50)`
`=2xx^(50)C_(1)xx100^(49)+2xx^(50)C_(3)xx100^(47)`
`+......+2xx^(50)C_(49)xx100-100^(50)`
`=2xx^(50)C_(3)xx100^(47)+....+2xx^(50)C_(49)xx100gt0`
`:.101^(50)-(100^(50)+99^(50)) gt 0`
`rArr " "101^(50)gt 100^(50)+99^(50)`