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)`

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

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

If p=(8+3sqrt(7))^n a n df=p-[p],w h e r e[dot] denotes the greatest integer function, then the value of p(1-f) is equal to a.1 b. 2 c. 2^n d. 2^(2n)

If R = (sqrt(2) + 1)^(2n+1) and f = R - [R] , where [ ] denote the greatest integer function, then [R] equal

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) .

Statement 1: If p is a prime number (p!=2), then [(2+sqrt(5))^p]-2^(p+1) is always divisible by p(w h e r e[dot] denotes the greatest integer function). Statement 2: if n prime, then ^n C_1,^n C_2,^n C_2 ,^n C_(n-1) must be divisible by ndot

Let S=sum_(r=1)^(117)(1)/(2[sqrt(r)]+1) where [.) denotes the greatest integer function.The value of S is

Evaluate: int_(-100)^(100)[tan^(-1)x]dx ,w h e r e[x] represents greatest integer function.

Let vecf(t)=[t] hat i+(t-[t]) hat j+[t+1] hat k , w h e r e[dot] denotes the greatest integer function. Then the vectors vecf(5/4)a n df(t),0lttlti are(a) parallel to each other(b) perpendicular(c) inclined at cos^(-1)2 (sqrt(7(1-t^2))) (d)inclined at cos^(-1)((8+t)/sqrt (1+t^2));