Question

$1+\frac{2.\cdot 1}{3.2}+\frac{\mathrm{2.5.1}}{{\mathrm{3.6.2}}^2}+\frac{\mathrm{2.5.8}}{36.9}\frac{1}{2^3}+---+{\infty }_{=}$

• A. $4^{\frac{1}{3}}$ (nearly)
• B. $3^{\frac{1}{4}}$
• C. $\sqrt{3}$
• D. $\sqrt{3}-1$

Answer 1

The correct option is A $4^{\frac{1}{3}}$ (nearly)

${(1-x)}^{-p/q}=1+\frac{p}{q}\left(x\right)+\frac{p}{q}(1+\frac{p}{q})\frac{x^2}{1.2}+\frac{p}{q}(1+\frac{p}{q})(2+\frac{p}{q})\frac{x^3}{\mathrm{1.2.3}}+...$

$|x|<1+\left(\frac{p}{q}\right)...(n+\frac{p}{q})\frac{x^n}{\mathrm{1.2.3...}n}+...\infty$

Comparing given summation with above expansion

$\frac{p}{q}=\frac{2}{3},x=\frac{-1}{2}$

${(1-\frac{1}{2})}^{-2/3}={\left(\frac{1}{2}\right)}^{-2/3}=2^{2/3}=4^{1/3}$

$=1+\left(\frac{2}{3}\right).\left(\frac{1}{2}\right)+\left(\frac{2}{3}\right)\left(\frac{5}{3}\right)\times {\left(\frac{1}{2}\right)}^2\frac{1}{1.2}+\left(\frac{2}{3}\right)\left(\frac{5}{3}\right)\left(\frac{8}{3}\right)\times {\left(\frac{1}{2}\right)}^3\frac{1}{\mathrm{1.2.3}}$