Use the following expansion to find the following (x- 3/x^2)^9 containing a term x^6.
We'll use the formula of general term of binomial expansion (a+b)^n:
T(k+1) = C(n,k)*[a^(n-k)]*b^k
Let a = x, b = -3/x^2, n = 9
T(k+1) = C(9,k)*[x^(9-k)]*[(-3/x^2)^k]
This term must contain x^6, therefore we'll consider just the power of x.
We'll use the negative power property:
(1/x^2)^k = x^(-2k)
x^(9-k)*x^(-2k) = x^6
We'll add the exponents from the left side:
x^(9-k-2k) = x^6
Since the bases both sides are matching, we'll equate the exponents:
9-3k = 6
We'll divide by 3:
3 - k = 2
We'll keep k to the left side and we'll subtract 3 both sides:
-k = -3 + 2
-k = -1
k = 1
Since k = 1, therefore the second term of the expansion, T2, will contain x^6.