Asked by steve
which term in the expansion of (1/(3x^2)-x)^15 is the constant term?
A. 10th
B. 11th
C. 12th
D. 13th
A. 10th
B. 11th
C. 12th
D. 13th
Answers
Answered by
MathMate
The constant term is the term where the power of x in the product (1/(3x²))^r*x^(15-r) is zero, i.e.
(x^(-2))^r*x^(15-r)=x^0
=>
-2r+(15-r)=0
r=5
In general, a polynomial in x is written in descending order of the power of x.
So the powers of x for the expansion are:
x^0*(x^(-2))^15 = x^(-30)
...
x^10*(x^(-2))^5 = x^0
...
x^12*(x^(-2))^3 = x^6
x^13*(x^(-2))^2 = x^9
x^14*(x^(-2))^1 = x^12
x^15*(x^(-2))^0 = x^15
Can you figure out which term gives the constant term?
(x^(-2))^r*x^(15-r)=x^0
=>
-2r+(15-r)=0
r=5
In general, a polynomial in x is written in descending order of the power of x.
So the powers of x for the expansion are:
x^0*(x^(-2))^15 = x^(-30)
...
x^10*(x^(-2))^5 = x^0
...
x^12*(x^(-2))^3 = x^6
x^13*(x^(-2))^2 = x^9
x^14*(x^(-2))^1 = x^12
x^15*(x^(-2))^0 = x^15
Can you figure out which term gives the constant term?
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