The sum of all the coefficients in the binomial expansion of ((1+2 x)^n ) is 6561. Let (R=(I+2 x)^n=I+F ), where (I N ) and (0 < F < l ). If (x= 1 2 ), then (1- F 1+( 2 -1)^4 = )

Step-by-step solution

It is given that sum of all the coefficients in the binomial expansion of ((1+2 x)^n ) is (6561=(1+2)^n ) on putting (x=1 ). ( array ll & 3^n=6561 & n=8 array ) Now, at (x= 1 2 ), then (R=(1+2 x)^n=I+F ) ( aligned & R=( 2 +1)^8=I+F, where I N and 0 < F < 1 & ( 2 -1)^8=F^ , where 0 < F^ < 1 & ( 2 +1)^8+( 2 -1)^8=I+ (F+F^ ) & 2 [( 2 )^8+ ^8 C₂( 2 )^6+ ^8 C₄( 2 )^4+ ^8 C₆( 2 )^2+ ^8 C₈ ] & =I+ (F+F^ ) & Even integer =I+ (F+F^ ) & F+F^ Integer & 0 < F < 1 and 0 < F^ < 1 0 < F+F^ < 2 & So, F+F^ =1 & F=1-F^ =1-( 2 -1)^8