(Z1+Z2)

^{n}=∑n!(k!(n-k)!) Z1

^{k}Z2

^{(n-k)}where the sum runs from k=0, to n. In our case Z1=-2x

^{6}and Z2=(5/x) so we can see that each term in the sum will look like

{ n!/(k!(n-k)!)} *(-2x

^{6k})*(5/x)

^{(21-k)}

now the constant term is the term that does not have an "x" in it. The only term that will not have an x in it is the where 6k-21+k=0 or 7k=21 and k=3. In that term the x's will have the same exponent and will therefore cancel leaving -2*5*{ 21!/(3!(21-3)!)} as the constant term which is -10*1330=-13,300.

Regards

Jim

Dalia S.

04/01/14