I hope this isn't to late to help you but this problem can be treated as a binomial expansion i.e.
(Z1+Z2)n=∑n!(k!(n-k)!) Z1kZ2(n-k) where the sum runs from k=0, to n. In our case Z1=-2x6 and Z2=(5/x) so we can see that each term in the sum will look like
{ n!/(k!(n-k)!)} *(-2x6k)*(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
(Z1+Z2)n=∑n!(k!(n-k)!) Z1kZ2(n-k) where the sum runs from k=0, to n. In our case Z1=-2x6 and Z2=(5/x) so we can see that each term in the sum will look like
{ n!/(k!(n-k)!)} *(-2x6k)*(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