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