Q58 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
$1-\frac{3}{16}+\frac{1 \cdot 4}{1 \cdot 2}\left(\frac{3}{16}\right)^2-\frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3}\left(\frac{3}{16}\right)^3+\ldots$
- A$\left(\frac{15}{6}\right)^{3 / 8}$
- Correct answerB$\left(\frac{4}{5}\right)^{2 / 3}$
- C$\left(\frac{7}{4}\right)^{1 / 16}$
- D$\left(\frac{4}{15}\right)^{-2 / 5}$
Solution
We have,
$1-\frac{3}{16}+\frac{1 \cdot 4}{1 \cdot 2}\left(\frac{3}{16}\right)^2-\frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3}\left(\frac{3}{16}\right)^3+\ldots=$
We know that,
$(1-x)^n=1-n x+\frac{n(n-1)}{2 !} x^2-\frac{n(n-1)(n-2)}{3 !} x^3$
Let $(1-x)^n=1-\frac{3}{16}+\frac{1 \cdot 4}{1 \cdot 2}\left(\frac{3}{16}\right)^2-\frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3}\left(\frac{3}{16}\right)^3 \ldots$
$\therefore \quad n x=\frac{3}{16}$ and $\frac{n(n-1)}{2 !} x^2=\frac{1 \cdot 4}{2 !}\left(\frac{3}{16}\right)^2$
$n x=\frac{3}{16}$ and $n(n-1) x^2=4\left(\frac{3}{16}\right)^2$
$n^2 \cdot x^2=\frac{3^2}{16}$ and $n(n-1) x^2=4\left(\frac{3}{16}\right)^2$
$n(n-1) x^2=4 n^2 x^2$
$n^2-n=4 n^2$
$n-1=4 n \Rightarrow n=-\frac{1}{3}$
$x=\frac{3}{16} \times-3=\frac{-9}{16}$
$\because \quad(1-x)^n=\left(1+\frac{9}{16}\right)^{-1 / 3}=\left(\frac{25}{16}\right)^{-1 / 3}=\left(\frac{4}{5}\right)^{2 / 3}$
Hence, the correct answer is option: 'B'.