If I_n= _ 0^ 4 ^n x d x , for n 2 , then I_n+I_ n-2 =

Step-by-step solution

Given I_n = ₀^ 4 ^ n x dx . Consider the sum I_n + I_ n-2 = ₀^ 4 ^ n x dx + ₀^ 4 ^ n-2 x dx . I_n + I_ n-2 = ₀^ 4 ^ n-2 x ( ² x + 1) dx . Using the identity ² x + 1 = ² x , we have: I_n + I_ n-2 = ₀^ 4 ^ n-2 x ² x dx . Let u = x , then du = ² x dx . When x = 0 , u = 0 . When x = 4 , u = 1 . I_n + I_ n-2 = ₀¹ u^ n-2 du . I_n + I_ n-2 = [ u^ n-1 n-1 ]₀¹ = 1 n-1 . Answer: 1 n-1