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

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

Let R=(5sqrt(5)+11)^(31)=1+f, where I is an integer and f is the fractional part of R then Rf is equal to

Statement-1: The integeral part of (8+3sqrt(7))^(20) is even. Statement-2: The sum of the last eight coefficients in the expansion of (1+x)^(16) is 2^(15) . Statement-3: if R(5sqrt(5)+11)^(2n+1)=[R]+F , where [R] denotes the greatest integer in R, then RF=2^(2n+1) .

Let R=(5sqrt(5)+11)^(2n+1) and f=R-[R] where [1 denotes the greatest integer function,prove that Rf=4^(n+1)

Consider the binomial expansion of R = (1 + 2x )^(n) = I + f , where I is the integral part of R and f is the fractional part of R , n in N . Also , the sum of coefficient of R is 2187. The value of (n+ Rf ) "for x" = (1)/(sqrt(2)) is

Let R=(2+sqrt(3))^(2n) and f=R-[R] where [ denotes the greatest integer function,then R(1-f) is equal to

If t_r is the rth term is the expansion of (1+a)^n , in ascending power of a , prove that r (r +1) t_(r+2) = (n - r + 1) (n - r) a^2 t_r

Let R = (sqrt2+1)^(2n+1),ninN and f= R- [R], where [] denote the greatest integer function, Rf is equal to

If the coefficients of a^(r-1),a^(r) and in the expansion of in the expansion of are in arithmetic progression,prove that n^(2)-n(4r+1)+4r^(2)-2=0