f(x) = 1/(3x^2 - 2)
I'm correcting a test and I got this one wrong. I think I made it a lot more difficult than it really is. I was finding the derivatives and plugging in zero but I think this is just a geometric series so is this the right way to do it?
f(x) = 1/(3x^2 - 2) = (-1/2)(1/(1 - (3/2)x^2))
(-1/2) ∑ ((3x^2)/2)^n
= ∑ (-(3)^n(x^2n))/(2^(n + 1)) <----- is that the correct answer??
I'm correcting a test and I got this one wrong. I think I made it a lot more difficult than it really is. I was finding the derivatives and plugging in zero but I think this is just a geometric series so is this the right way to do it?
f(x) = 1/(3x^2 - 2) = (-1/2)(1/(1 - (3/2)x^2))
(-1/2) ∑ ((3x^2)/2)^n
= ∑ (-(3)^n(x^2n))/(2^(n + 1)) <----- is that the correct answer??
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Yes, that answer is indeed correct.