If we count the first term as the k=0th term, then the kth term of (A+B)n is (nCk)An-kBk The 11th term of the expansion of (x+2y)14 is the k=10th term if we count the first term as the 0th term. So we substitute k=10, A=x, B=2y, n=14. (nCk)An-kBk (14C10)x14-10(2y)10 1001x4(2y)10 1001x4210y10 1001x41024y10 1025024x4y10 Edwin