If $x=\frac{2}{5}+\frac{1 \cdot 3}{2 !}\left(\frac{2}{5}\right)^2+\frac{1 \cdot 3 \cdot 5}{3 !}\left(\frac{2}{5}\right)^3+\ldots$, then $x+\frac{1}{x}=$

Question

Aicharya · Accepted Answer

Let $x=(1+y)^n-1=n y$ $\begin{aligned} + & \frac{n(n-1)}{2 !} y^2+\frac{n(n-1)(n-2)}{3 !} y^3+\ldots \ldots \ldots . . \ & =\frac{2}{5}+\frac{1.3}{2 !}\left(\frac{2}{5}\right)^2+\frac{1 \cdot 3 \cdot 5}{3 !}\left(\frac{2}{5}\right)^3+\ldots \ldots \ldots . . . \end{aligned}$ On comparing first three terms, we get $n y=\frac{2}{5}, \frac{n(n-1)}{2 !} y^2=\frac{1 \cdot 3}{2 !}\left(\frac{2}{5}\right)^2$ and $\quad \frac{n(n-1)(n-2)}{3 !} y^3=\frac{1 \cdot 3 \cdot 5}{3 !}\left(\frac{2}{5}\right)^3$ From first two relations, we get $\begin{array}{rlrl} & & \frac{n y(n y-y)}{2 !} & =\frac{\frac{2}{5}\left(\frac{2}{5}-y\right)}{2 !}=\frac{1 \cdot 3}{2 !}\left(\frac{2}{5}\right)^2 \ \Rightarrow \quad & \left(\frac{2}{5}\right)-y & =1 \cdot 3\left(\frac{2}{5}\right) \ \Rightarrow & & y & =\frac{2}{5}-\frac{6}{5}=-\frac{4}{5} \text { and } n=-\frac{1}{2} \end{array}$ Now, on putting $n=-\frac{1}{2}$ and $y=-\frac{4}{5}$ in LHS of third relation, we get $\begin{aligned} & =\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3 !}\left(-\frac{4}{5}\right)^3 \ & =\frac{\frac{1}{2}\left(\frac{1+2}{2}\right)\left(\frac{1+4}{2}\right)}{3 !} 2^3\left(\frac{2}{5}\right)^3=\frac{1 \cdot 3 \cdot 5}{3 !}\left(\frac{2}{5}\right)^3=\text { RHS } \end{aligned}$ So, $x=\left(1-\frac{4}{5}\right)^{-\frac{1}{2}}-1=\sqrt{5}-1$ so, $\frac{1}{x}=\frac{\sqrt{5}+1}{4}$ So, $x+\frac{1}{x}=\sqrt{5}-1+\frac{\sqrt{5}+1}{4}=\frac{5 \sqrt{5}-3}{4}$ Hence, option (d) is correct.

If \(x=\frac{2}{5}+\frac{1 \cdot 3}{2 !}\left(\frac{2}{5}\right)^2+\frac{1 \cdot 3 \cdot 5}{3…

Solution