Question

If A and B are the coefficients of $- \frac{2}{\sqrt{3}} < x < \frac{2}{\sqrt{3}}$ in the expansions of $(a + bx)^{- 2} = \frac{1}{4} - 3x + ......$ and $(a,b)$ respectively, then.

• A. $6^{1/3}\left\lbrack 1 + \frac{x}{6} + \frac{2x^{2}}{6^{2}} + .... \right\rbrack$
• B. $6^{- 1/3}\left\lbrack 1 + \frac{x}{6} + \frac{2x^{2}}{6^{2}} + .... \right\rbrack$
• C. $6^{1/3}\left\lbrack 1 - \frac{x}{6} + \frac{2x^{2}}{6^{2}} - .... \right\rbrack$
• D. None of these

Answer 1

Answer: (B)

$6^{- 1/3}\left\lbrack 1 + \frac{x}{6} + \frac{2x^{2}}{6^{2}} + .... \right\rbrack$

Answer 2

$\therefore 3^{180} = (3^{4})^{45}$

$3^{183} = 3^{180}3^{3}$

$183! + 3^{183}$

⇒ $T_{r + 1} = {_{}^{256}C}_{r}(\sqrt{3})^{256 - r}(\sqrt[8]{5})^{r}$ ⇒ $= {_{}^{256}C}_{r}(3)^{\frac{256 - r}{2}}(5)^{r/8}$.