The coefficient of x^3 in the power series ex | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We use the generalized binomial expansion: $(1+z)^{-n} = 1 - nz + \frac{n(n+1)}{2!}z^2 - \frac{n(n+1)(n+2)}{3!}z^3 + \dots$For $(1+3x)^{-3}$,we ha

Vedclass · Accepted Answer

-$27$

Vedclass · Answer

(A) We use the generalized binomial expansion: $(1+z)^{-n} = 1 - nz + \frac{n(n+1)}{2!}z^2 - \frac{n(n+1)(n+2)}{3!}z^3 + \dots$For $(1+3x)^{-3}$,we have $n=3$ and $z=3x$:$(1+3x)^{-3} = 1 - 3(3x) + \frac{3(4)}{2}(3x)^2 - \frac{3(4)(5)}{6}(3x)^3 + \dots$$= 1 - 9x + 54x^2 - 270x^3 + \dots$Now,multiply by $(1+4x-3x^2)$:$(1+4x-3x^2)(1-9x+54x^2-270x^3 + \dots)$The coefficient of $x^3$ is obtained by:$1(-270) + 4(54) - 3(-9) = -270 + 216 + 27 = -27$.

The coefficient of $x^3$ in the power series expansion of $\frac{1+4x-3x^2}{(1+3x)^3}$ is

Similar Questions

The sum of the series $\frac{3}{4 \cdot 8}-\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}-\ldots$

$\frac{1}{4}-\frac{5}{4 \cdot 8}+\frac{5 \cdot 9}{4 \cdot 8 \cdot 12}-\ldots=$

If $(a+bx)^{-3} = \frac{1}{27} + \frac{1}{3}x + \dots$,then the ordered pair $(a, b)$ is equal to

If $(a + bx)^{-2} = \frac{1}{4} - 3x + \dots$,then $(a, b) = $

The coefficient of $x^2$ in the expansion of $(1+x)^2(8-x)^{-\frac{1}{3}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module