frac{1}{8} - frac{7}{8 times 12} + frac{7 tim | vedclass

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

Question

(B) Let $S = \frac{1}{8} - \frac{7}{8 	imes 12} + \frac{7 	imes 10}{8 	imes 12 	imes 16} - \ldots$ Using the binomial expansion $(1+x)^n = 1 + nx

Vedclass · Accepted Answer

$\sqrt[3]{\frac{4}{7}} - \frac{3}{4}$

Vedclass · Answer

(B) Let $S = \frac{1}{8} - \frac{7}{8 	imes 12} + \frac{7 	imes 10}{8 	imes 12 	imes 16} - \ldots$ Using the binomial expansion $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots$ Consider $(1 + \frac{3}{4})^{-1/3} = 1 + (-\frac{1}{3})(\frac{3}{4}) + \frac{(-\frac{1}{3})(-\frac{4}{3})}{2!}(\frac{3}{4})^2 + \frac{(-\frac{1}{3})(-\frac{4}{3})(-\frac{7}{3})}{3!}(\frac{3}{4})^3 + \ldots$ $= 1 - \frac{1}{4} + \frac{4}{18} 	imes \frac{9}{16} - \frac{28}{162} 	imes \frac{27}{64} + \ldots$ $= 1 - \frac{1}{4} + \frac{1}{8} - \frac{7}{8 	imes 12} + \ldots$ $= \frac{3}{4} + S$ Since $(1 + \frac{3}{4})^{-1/3} = (\frac{7}{4})^{-1/3} = \sqrt[3]{\frac{4}{7}}$,we have $\sqrt[3]{\frac{4}{7}} = \frac{3}{4} + S$ Therefore,$S = \sqrt[3]{\frac{4}{7}} - \frac{3}{4}$

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

Similar Questions

If $x$ is a positive real number and the first negative term in the expansion of $(1+x)^{27/5}$ is $t_k$,then $k=$

If $x=1+\frac{3}{1!} \times \frac{1}{6}+\frac{3 \times 7}{2!}\left(\frac{1}{6}\right)^2+\frac{3 \times 7 \times 11}{3!}\left(\frac{1}{6}\right)^3+\ldots$,then $x^4$ equals

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

If $3x = 1 + \frac{5}{8} + \frac{5 \times 9}{8 \times 16} + \frac{5 \times 9 \times 13}{8 \times 16 \times 24} + \dots$,then $x^4 + 4x^3 + 6x^2 + 4x = $

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