Consider the expansion of (x-1/6x^2)^9. Find the constant term in this expansion.
the general term is ...
term(r+1) = C(9,r) x^r ((-1/6)x^2)^(9-r)
= C(9,r) (-1/6)^(9-r) x^r x^(18-2r)
= C(9,r) (-1/6)^(9-r) x^(18-r)
You will have a constant term when it contains no variable x, that is, when 18-r = 0
so r = 18
But that makes no sense since 0 ≤ r ≤ 9
confirmed by Wolfram, notice every term contains an x term
https://www.wolframalpha.com/input/?i=expand+%28x-1%2F6x%5E2%29%5E9
check your question, fix it, then follow my steps
or, maybe
term(r+1) = C(9,r) x^r (-1/(6x^2))^(9-r)
= C(9,r) (-1/6)^(9-r) x^r x^(2r-18)
= C(9,r) (-1/6)^(9-r) x^(3r-18)
r = 6
C(9,6) (-1/6)^3 = -7/18
To find the constant term in the expansion of (x - 1/6x^2)^9, we need to determine the term that does not contain any powers of x.
Let's use the binomial theorem to expand this expression:
(x - 1/6x^2)^9 = C(9,0)*(x)^9*(-1/6x^2)^0 + C(9,1)*(x)^8*(-1/6x^2)^1 + ... + C(9,9)*(x)^0*(-1/6x^2)^9
Here, C(n, r) represents the binomial coefficient, which is given by n! / (r!(n-r)!), with n and r being the numbers of trials and successes, respectively.
For the constant term, we need to find the term where x and (1/6x^2) cancel each other out. This happens when the power of x is 0.
So, we are looking for the term: C(9, m)*(x)^(9-m)*(-1/6x^2)^m, where m is the power of (-1/6x^2).
For the constant term, we want (x)^(9-m)*(-1/6x^2)^m = 1.
Setting x^(9-m) = 1 and (-1/6x^2)^m = 1, we can solve for m.
x^(9-m) = 1 => 9 - m = 0 => m = 9
(-1/6x^2)^m = 1 => (-1/6x^2)^9 = 1
So, the constant term comes from the term: C(9, 9)*(x)^0*(-1/6x^2)^9
Calculating this term:
C(9, 9) = 1
x^0 = 1
(-1/6x^2)^9 = (-1/6)^9 * (x^(-2))^9 = (-1/6)^9 * x^(-18) = 1/6^9 * x^(-18)
Therefore, the constant term in the expansion is 1/6^9.