If |x|

Question

(D) We have the expression $\frac{1+2x}{(1-2x)^2} = (1+2x)(1-2x)^{-2}$.Using the binomial expansion for negative indices,$(1-y)^{-2} = \sum_{k=0}^{\in

Vedclass · Accepted Answer

$(2r+1) 2^r$

Vedclass · Answer

(D) We have the expression $\frac{1+2x}{(1-2x)^2} = (1+2x)(1-2x)^{-2}$.Using the binomial expansion for negative indices,$(1-y)^{-2} = \sum_{k=0}^{\infty} (k+1)y^k$.Substituting $y = 2x$,we get $(1-2x)^{-2} = \sum_{k=0}^{\infty} (k+1)(2x)^k = \sum_{k=0}^{\infty} (k+1) 2^k x^k$.Now,multiply by $(1+2x)$:$(1+2x) \sum_{k=0}^{\infty} (k+1) 2^k x^k = \sum_{k=0}^{\infty} (k+1) 2^k x^k + \sum_{k=0}^{\infty} 2(k+1) 2^k x^{k+1}$.To find the coefficient of $x^r$,we take the term where $k=r$ from the first sum and the term where $k+1=r$ (i.e.,$k=r-1$) from the second sum:Coefficient of $x^r = (r+1) 2^r + 2((r-1)+1) 2^{r-1}$.$= (r+1) 2^r + 2(r) 2^{r-1} = (r+1) 2^r + r 2^r$.$= (r+1+r) 2^r = (2r+1) 2^r$.

If $|x| < \frac{1}{2}$,then the coefficient of $x^r$ in the expansion of $\frac{1+2x}{(1-2x)^2}$ is

Similar Questions

If $|x| < 1$,then the number of terms in the expansion of $[\frac{1}{2}(1 \cdot 2 + 2 \cdot 3 x + 3 \cdot 4 x^2 + . . . . . . \infty)]^{-25}$ is

The sum of the series $1 + \frac{1 \times 3}{6} + \frac{1 \times 3 \times 5}{6 \times 8} + \dots \infty$ is

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

$1+\frac{4}{15}+\frac{4 \times 10}{15 \times 30}+\frac{4 \times 10 \times 16}{15 \times 30 \times 45}+\ldots \quad \infty=$

Assertion $(A)$: If $|x| < 1$,then $\sum_{n=0}^{\infty}(-1)^n x^{n+1} = \frac{x}{x+1}$.
Reason $(R)$: If $|x| < 1$,then $(1+x)^{-1} = 1-x+x^2-x^3+\dots$.
Which one of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module