The coefficient of x^2 in the expansion of (1 | vedclass

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

Question

(B) Using the binomial expansion $(1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \dots$,we expand the two terms:$(1-3x)^{\frac{1}{3}} = 1 + \frac{1}{3}(-3x

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(B) Using the binomial expansion $(1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \dots$,we expand the two terms:$(1-3x)^{\frac{1}{3}} = 1 + \frac{1}{3}(-3x) + \frac{\frac{1}{3}(\frac{1}{3}-1)}{2!}(-3x)^2 + \dots = 1 - x - x^2 + \dots$$(1+2x)^{-\frac{1}{2}} = 1 + (-\frac{1}{2})(2x) + \frac{(-\frac{1}{2})(-\frac{1}{2}-1)}{2!}(2x)^2 + \dots = 1 - x + \frac{3}{2}x^2 + \dots$Multiplying these expansions: $(1 - x - x^2 + \dots)(1 - x + \frac{3}{2}x^2 + \dots)$The coefficient of $x^2$ is obtained by: $(1 	imes \frac{3}{2}) + (-1 	imes -1) + (-1 	imes 1) = \frac{3}{2} + 1 - 1 = \frac{3}{2}$.

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

Similar Questions

If $x$ is so small that $x^2$ and higher powers of $x$ can be neglected,then the approximate value of $\left(1+\frac{3}{4} x\right)^{\frac{1}{2}}\left(1-\frac{2 x}{3}\right)^{-2}$ is

To expand $(1 + 2x)^{-1/2}$ as an infinite series,the range of $x$ should be

The coefficient of ${x^n}$ in the expansion of ${(1 - 9x + 20{x^2})^{-1}}$ is

The coefficient of $x^r$ in the expansion of $\frac{1}{\sqrt[3]{(1-2 x)^2}}$ is

If $y = \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \dots \infty$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module