Q54 / 90 · KVPY 2012 (SB/SX) · Mathematics
Define a sequence $\left(a_{n}\right)$ by $a_{1}=5, a_{n}=a_{1} a_{2} \ldots a_{n-1}+4$ for $n>1$. Then $\lim _{n \rightarrow \infty} \frac{\sqrt{a_{n}}}{a_{n-1}}$
- AEquals $1 / 2$
- Bequals 1
- Correct answerCequals $2 / 5$
- Ddoes not exist
Solution
$a_{1}=5$
$a_{n}=a_{1} a_{2} \ldots \ldots a_{n-1}+4$
$a_{2}=a_{1}+4=9$
$a_{3}=a_{1} \cdot a_{2}+4=5 \times 9+4=49$
$a_{4}=a_{1} a_{2} a_{3}+4=2209$
$a_{5}=a_{1} a_{2} a_{3} a_{4}+4=4870849=(2207)^{2}$
$a_{5}=\left(a_{4}-2\right)^{2}$
$a_{4}=(49-2)^{2}=\left(a_{3}-2\right)^{2}$
$a_{3}=(9-2)^{2}=\left(a_{2}-2\right)^{2}$
$a n=\left(a_{n-1}-2\right)^{2}$
$\sqrt{a_{n}}=a_{n-1}-2$
$\frac{\sqrt{a_{n}}}{a_{n-1}}=\frac{a_{n-1}-2}{a_{n-1}}=1-\lim _{n \rightarrow \infty} \frac{2}{a_{n-1}}$
$\because \lim _{n \rightarrow \infty} a_{n-1}=\infty$
$=1$
Hence, the correct answer is option: 'C'.