Question 1060088
<pre><b>
If we count the first term as the k=0th term, then the
kth term of (A+B)<sup>n</sup> is (nCk)A<sup>n-k</sup>B<sup>k</sup>

The 11th term of the expansion of (x+2y)<sup>14</sup> 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)A<sup>n-k</sup>B<sup>k</sup> 

(14C10)x<sup>14-10</sup>(2y)<sup>10</sup>

1001x<sup>4</sup>(2y)<sup>10</sup>

1001x<sup>4</sup>2<sup>10</sup>y<sup>10</sup>

1001x<sup>4</sup>1024</sup>y<sup>10</sup>

1025024x<sup>4</sup>y<sup>10</sup>

Edwin</pre></b>