Q75 / 110 · VITEEE 2024 · Mathematics
The sum of $1+\frac{1}{4}+\frac{1 \cdot 3}{4 \cdot 8}+\frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12}+\ldots \infty$ is
- A$\sqrt{3}$
- B$\frac{1}{\sqrt{2}}$
- Correct answerC$\sqrt{2}$
- D$2^{\frac{3}{2}}$
Solution
Comparing with $1+n x+\frac{n(n-1)}{2!} x^2+\ldots$
$n x=\frac{1}{4}$...(i)
and $\quad \frac{n(n-1)}{2!} x^2=\frac{1 \cdot 3}{4 \cdot 8}$
$\begin{aligned}
& \Rightarrow \quad \frac{n \times(n-1) x}{2!}=\frac{1 \cdot 3}{4 \cdot 8} \\
& \text { put : } n x=\frac{1}{4} \\
& \Rightarrow \quad \frac{1}{4} \frac{(n-1) x}{2!}=\frac{1 \cdot 3}{4 \cdot 8} \\
& \Rightarrow \quad \frac{(n-1) x}{2}=\frac{3}{8} \\
& \Rightarrow \quad \frac{n x}{2}-\frac{x}{2}=\frac{3}{8} \\
& \Rightarrow \quad \frac{1}{8}-\frac{x}{2}=\frac{3}{8} \\
& \Rightarrow \quad \frac{x}{2}=-\frac{2}{8} \\
& \Rightarrow \quad x=-\frac{1}{2}
\end{aligned}$
Putting the value of $x$ in Eq. (i),
$\begin{aligned}
n \times\left(-\frac{1}{2}\right) & =\frac{1}{4} \\
\Rightarrow \quad n & =-\frac{1}{2} \\
\therefore \quad \text { Sum of series } & =(1+x)^n=\left(1-\frac{1}{2}\right)^{-\frac{1}{2}} \\
& =\left(\frac{1}{2}\right)^{-\frac{1}{2}}=\sqrt{2}
\end{aligned}$
Hence, the correct answer is option: 'C'.