1+frac{1}{3}+frac{1 times 3}{3 times 6}+frac{ | vedclass

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

Question

(D) The given series is $S = 1 + \frac{1}{3} + \frac{1 	imes 3}{3 	imes 6} + \frac{1 	imes 3 	imes 5}{3 	imes 6 	imes 9} + \ldots \infty$. We ca

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(D) The given series is $S = 1 + \frac{1}{3} + \frac{1 	imes 3}{3 	imes 6} + \frac{1 	imes 3 	imes 5}{3 	imes 6 	imes 9} + \ldots \infty$. We can rewrite the general term by multiplying the denominator by $3^n n!$: $S = 1 + \frac{1}{3(1!)} + \frac{1 	imes 3}{3^2(2!)} + \frac{1 	imes 3 	imes 5}{3^3(3!)} + \ldots$. Comparing this with the binomial expansion $(1-x)^{-p/q} = 1 + \frac{p}{q}x + \frac{p(p+q)}{2!}(\frac{x}{q})^2 + \frac{p(p+q)(p+2q)}{3!}(\frac{x}{q})^3 + \ldots$. Here,$p=1, q=2$,and $\frac{x}{q} = \frac{1}{3} \Rightarrow x = \frac{2}{3}$. Thus,$S = (1 - \frac{2}{3})^{-1/2} = (\frac{1}{3})^{-1/2} = \sqrt{3}$.

$1+\frac{1}{3}+\frac{1 \times 3}{3 \times 6}+\frac{1 \times 3 \times 5}{3 \times 6 \times 9}+\ldots \text{ to } \infty =$

Similar Questions

If $x$ is positive,the first negative term in the expansion of $(1 + x)^{27/5}$ is

If the value of $x$ is so small that $x^2$ and higher powers can be neglected,then $\frac{\sqrt{1 + x} + \sqrt[3]{(1 - x)^2}}{1 + x + \sqrt{1 + x}}$ is equal to

If $(-c, c)$ is the set of all values of $x$ for which the expansion of $(7-5x)^{-2/3}$ is valid,then $5c + 7 =$

$\frac{1}{8} - \frac{7}{8 \times 12} + \frac{7 \times 10}{8 \times 12 \times 16} - \ldots =$

For $x>0$,if $p^{\text{th}}$ term is the first negative term in the expansion of $(1+\frac{3x}{5})^{22/3}$ and in the expansion of $(1-\frac{3x}{5})^{22/3}$ from $r^{\text{th}}$ term onwards all the terms are positive,then the number of terms in the expansion of $(px+\frac{r}{x})^{pr}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module