Let `(6sqrt(6)+14)^(2n+1)=R`, if f be the fractional part of R, then prove that `Rf=20^(2n+1)`

Options

A. `20^(n)`
B. `20^(2n)`
C. `20^(2n+1)`
D. none of these

Answer

c

Explanation

Let G = `(6 sqrt(6 - 14)^(2n +1)` . Then
`R- G = (6sqrt(6) + 14)^(2n+1) - (6sqrt(6) - 14 )^(2n+1)`
`rArr R-G = 2 {"^(2n+1)C_(1) (6 sqrt(6))^(2n) (14) +…}`
` rArr R-G ` = An even integer
`rArr [R] + f - G ` = An even integer
`rArr f = G`
`therefore Rf = RG = (6 sqrt(6) + 14)^(2n-1) (6sqrt(6) -14)^(2n+1) = 20^(2n +1)`.