discrete maths

Question

What is  the coefficient of $x^{12}$ in the power series of $\dfrac{x^3}{(1+2x)^2}?$

Comments

use binomial

$(x^3) \underbrace{(1+2x)^{-2}}$

find coefficient of  $(x^9)$ from 2nd term

 it will be : $\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8) \times (2x)^{9}}{9!}$

n = -2 here  

$\frac{-2(-3)(-4)(-5)(-6)(-7)(-8)(-9)(-10) \times (2x)^{9}}{9!}$

=$- \frac{2(3)(4)(5)(6)(7)(8)(9)(10) \times (2x)^{9}}{9\times 8!}$

= $10(-1) \times 512 x^{9} = -5120$

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but here n = 2 na

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why n=2   , i took  denominator above so it becomes -2

Answer 1

Hope This helps