For n 2 , let I_n= ₀^ / 4 ^n x d x and F_n=I_n+I_ n-2 . Then, F_n-F_ n+1 =

Step-by-step solution

I_n= ₀^ / 4 ^n x d x, n 2 and aligned F_n & =I_n+I_ n-2 & = ₀^ / 4 ( ^n x+ ^ n-2 x ) d x & = ₀^ / 4 ( ^2 x+1 ) ^ n-2 x d x & = ₀^ / 4 ^ n-2 x ^2 x d x aligned Let x=t ^2 x d x=d t Then, at x= 4 t=1 and x=0 t=0 So, aligned F_n & = ₀^1 t^ n-2 d t= . t^ n-1 n-1 |₀ ^1 & = 1 n-1 (1)^ n-1 = 1 n-1 F_ n+1 & = 1 n aligned F_ n+1 = 1 n So, F_n-F_ n+1 = 1 n-1 - 1 n = n-(n-1) n(n-1) = 1 n(n-1)