In the binomial expansion of (1+x)^n , let C_k denote the coefficient of x^k . Three terms are defined as X = C_ r-1 C_ r-1 + C_r , Y = C_r C_r + C_ r+1 , and Z = C_ r+1 C_ r+1 + C_ r+2 . If X = 1 4 and Z = 3 8 , then the value of n is

Step-by-step solution

First, simplify the expression for X using Pascal's identity C_ r-1 + C_r = ,^ n C_ r-1 + ,^ n C_ r = ,^ n+1 C_ r : X = ,^ n C_ r-1 ,^ n+1 C_ r = n! (r-1)!(n-r+1)! (n+1)! r!(n-r+1)! = r n+1 Similarly, simplifying Y and Z gives: Y = ,^ n C_ r ,^ n+1 C_ r+1 = r+1 n+1 Z = ,^ n C_ r+1 ,^ n+1 C_ r+2 = r+2 n+1 Observe that X, Y, and Z form an Arithmetic Progression with a common difference d = 1 n+1 . We are given X = 1 4 and Z = 3 8 . The difference between Z and X is 2d : Z - X = r+2 n+1 - r n+1 = 2 n+1 Substitute the