Q58 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Mathematics

$\frac{1}{8}-\frac{7}{8.12}+\frac{7.10}{8.12 .16}-\ldots=$

  • A
    $\sqrt[3]{\frac{4}{7}}$
  • Correct answer
    B
    $\sqrt[3]{\frac{4}{7}}-\frac{3}{4}$
  • C
    $\sqrt[3]{\frac{4}{7}}+\frac{3}{4}$
  • D
    $\sqrt[3]{\frac{7}{4}}-\frac{3}{4}$

Solution

$\begin{aligned} & \left(\frac{4}{7}\right)^{\frac{1}{3}}=\left(\frac{7}{4}\right)^{-\frac{1}{3}}=\left(1+\frac{3}{4}\right)^{\frac{-1}{3}} \\ & =1-\frac{1}{3} \times \frac{3}{4}+\frac{\left(-\frac{1}{3}\right)\left(-\frac{1}{3}-1\right)}{2 !}\left(\frac{3}{4}\right)^2+\frac{\left(-\frac{1}{3}\right)\left(-\frac{1}{3}-2\right)}{3 !}\left(\frac{3}{4}\right)^3 \\ & +\frac{\left(-\frac{1}{3}\right)\left(-\frac{1}{3}-1\right)\left(-\frac{1}{3}-2\right)\left(-\frac{1}{3}-3\right)}{4 !}\left(\frac{3}{4}\right)^4+\ldots \\ & =1-\frac{1}{4}+\frac{1}{3} \times \frac{4}{3} \times \frac{1}{2} \times \frac{3^2}{4^2}-\frac{1}{3} \times \frac{4}{3} \times \frac{7}{3} \times \frac{1}{6} \times \frac{3^3}{4^3} \\ & +\frac{1}{3} \times \frac{4}{3} \times \frac{7}{3} \times \frac{10}{3} \times \frac{1}{24} \times \frac{3^4}{4^4}+\ldots . \\ & =\frac{3}{4}+\frac{1}{8}-\frac{7}{8 \times 12}+\frac{7 \times 10}{8 \times 12 \times 16} \ldots \ldots . \\ & \Rightarrow \sqrt[3]{\frac{4}{7}}-\frac{3}{4}=\frac{1}{8}-\frac{7}{8 \times 12}+\frac{7 \times 10}{8 \times 12 \times 16}\end{aligned}$

Hence, the correct answer is option: 'B'.

Relevant Questions