Let I_n = ₀^ /2 2nx x dx for n N . Then the value of _ k=1 ¹⁵ 2 I_ k+1 + I_k is equal to

Step-by-step solution

Given I_n = ₀^ /2 2nx x dx Consider I_ k+1 + I_k = ₀^ /2 (2k+2)x + 2kx x dx Using the identity C + D = 2 ( C+D 2 ) ( C-D 2 ) , we get: I_ k+1 + I_k = ₀^ /2 2 (2k+1)x x x dx = 2 ₀^ /2 (2k+1)x dx = 2 [ - (2k+1)x 2k+1 ]₀^ /2 = 2 ( 0 - ( -1 2k+1 ) ) = 2 2k+1 Therefore, 2 I_ k+1 + I_k = 2k+1 Now, we need to evaluate the sum: _ k=1 ¹⁵ (2k+1) = 3 + 5 + 7 + + 31 This is an arithmetic progression with 15 terms, first term a = 3 , and last term l = 31 . Sum = 15 2 (3 + 31) = 15 2 (34) = 15 17 = 255 . Answer: 255