show that the expansion of (2x² -1/x)^9 has a constant term of 672.?

Question

well ok...but i don't get how u get the answer.

Answer 1

You will only get a constant term when the following occurs: (2x^2)^3 (-1/x)^6

The coefficient is found as follows:
(2^3)*(-1)^6 * C(9,6) = 8*1*84 = 672

Answer 2

(a+b)^n=sume(nCk*a^n-k*b^k) , for k=0 to n
also a=2x^2; b=-1/x; n=9
the generale term=9Ck(2x^2)^9-k*(-1/x)^k
X^(18-3k)=X^0 then 18-3k=0; k=6
The 7-thin term is 9C6*2^3=9*8*7*8/3*2=672.The answer is yes.

Answer 3

k+1st term

= (9Ck)(2x^2)^k(-1/x)^9-k
= (9Ck)(2x^2)^k(-x)^(k-9)
= (9Ck) 2^kx^(3k-9)

for it to be constant 3k-9 = 0 or k = 3

constant term (9C3)2^3
= 9*8*7/6*8 = 12*7*8 = 84*8 = 672
proved