Find the constant term

Question

Find the constant term in the expansion of 

(-2x⁶+(5/x))²¹

Answer 1

Dalia,

        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)!) Z₁^KZ₂^(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)!)} *(-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

 

Answer 2

If you multily the top and bottom of this expression by x you get a constant of five so the constant of the resulting expansion will be five to the 21.

 

Hope that helps!

Deanna