Most Important Selected Qs for JEE AdvancedMathematicsDefinite Integration
The value of _ k=1 ^ 6^k (3^k-2^k ) (3^ k+1 -2^ k+1 ) can be equal to:
Options
- A_ x 0 1- 2 x x^2
- B_ n ( 2 n 2 2 n 3 2 n (n-1) n )^ 1 / n
- C_ / 2 ^0 | 2 x| d x 2
- D2 2 ₀^ / 4 (1+ x) d x
Correct answer
C. _ / 2 ^0 | 2 x| d x 2
Step-by-step solution
Let 3^k=x2^k=y 6^k (3^k-2^k ) (3^ k+1 -2^ k+1 ) = x y (x-y)(3 x-2 y) aligned & = y(3 x-2 y)-2 y(x-y) (x-y)(3 x-2 y) = y x-y - 2 y 3 x-2 y & = 2^k 3^k-2^k - 2^ k+1 3^ k+1 -2^ k+1 aligned _ k=1 ^ 6^k (3^k-2^k ) (3^ k+1 -2^ k+1 ) = _ k=1 ^ 2^k 3^k-2^k - 2^ k+1 3^ k+1 -2^ k+1 aligned & T₁= 2 3-2 - 2^2 3^2-2^2 & T₂= 2^2 3^2-2^2 - 2^3 3^3-2^3 & & T_k= 2^k 3^k-2^k - 2^ k+1 3^ k+1 -2^ k+1 aligned _ k=1 ^ T_k= _ k (2- 2 2^k 3 3^k-2 2^k )=2