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

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)a n df=R-[R]w h e r e[] denotes the greatest integer function, prove that Rf=4^(2n+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

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. If ith term is the geratest term for x= 1/3, then i equal

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. If kth term is having greatest coefficient , the sum of all possible value of k, is

If (6 sqrt6+14)^(2n+1)=R and F=[R] , where [R] denotes the greatest integer less than or equal to R thwn RF=

If R = (7 + 4 sqrt(3))^(2n) = 1 + f , where I in N and 0 lt f lt 1 , then R (1 - f) equals

If the coefficients of a^(r-1),\ a^r and \ a^(r+1) in the binomial expansion of (1+a)^n are in A.P., prove that n^2-n(4r+1)+4r^2-2=0.

If for positive integers r gt 1, n gt 2 , the coefficients of the (3r) th and (r+2) th powers of x in the expansion of (1+x)^(2n) are equal, then prove that n=2r+1 .

Let f (n)=(4n + sqrt(4n ^(2) -1))/( sqrt(2n +1 )+sqrt(2n-1)),n in N then the remainder when f (1) + f (2) + f (3) + ..... + f (60) is divided by 9 is.