Q63 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics
If $y=\frac{3}{4}+\frac{3.5}{4.8}+\frac{3.5 .7}{4.8 .12}+\ldots$ to $\infty$, then
- A$y^2-2 y+5=0$
- Correct answerB$y^2+2 y-7=0$
- C$y^2-3 y+4=0$
- D$y^2+4 y-6=0$
Solution
$y=\frac{3}{4}+\frac{3.5}{4.8}+\frac{3.5 .7}{4.8 .12}+\ldots \infty$
$$
\begin{aligned}
& y=\frac{3}{4.1 !}+\frac{3.5}{2 !}\left(\frac{1}{4}\right)^2+\frac{3.5 .7}{3 !}\left(\frac{1}{4}\right)^3+\ldots \infty \\
& y+1=1+\frac{3}{1 !}\left(\frac{1}{4}\right)^1+\frac{3.5}{2 !}\left(\frac{1}{4}\right)^2+\frac{3.5 .7}{3 !}\left(\frac{1}{4}\right)^3 \ldots \infty
\end{aligned}
$$
Comparing with
$$
\begin{aligned}
& 1+a\left(\frac{x}{b}\right)^1+\frac{a(a+b)}{2 !}\left(\frac{x}{b}\right)^2+\ldots . \infty \\
& a=3, b=2, \frac{x}{b}=\frac{1}{4} \\
& x=\frac{1}{4} \times 2=\frac{1}{2} \\
& y+1=\left(1-\frac{1}{2}\right)^{-\frac{3}{2}}=\left(\frac{1}{2}\right)^{-\frac{3}{2}} \\
& (y+1)^2=2^3 \\
& \Rightarrow y^2+2 y-7=0
\end{aligned}
$$
Hence, the correct answer is option: 'B'.