help in binomial question

Question

if in the expansion of (2^x+1/4^x)^n, T3/T2=7 and sum of coefficient of 2nd and 3rd term is 36 the value of x is?

Answer 1

Second term coefficient is n; 3rd term coefficient is n(n-1)/2. The sum is n+n^2/2-n/2=n/2+n^2/2=36, so n^2+n-72=0 and (n+9)(n-8)=0, so since n assumed to be positive, n=8. (a+b)^8=a^8+8a^7b+28a^6b^2+...

Put a=2^x and b=1/4^x=4^-x=2^-2x. Expansion is 2^8x+8*2^(7x-2x)+28*2^(6x-4x)+...=2^8x+2^(5x+3)+7*2^(2x+2)+...

T3/T2=7 so 7*2^(2x+2)/2^(5x+3)=7; 2^(2x+2-5x-3)=1; -3x-1=0 (2^0=1), x=-1/3.