Sequence And Series Question 173
Question: If $ f(x) $ is a function satisfying $ f(x+y)=f(x)f(y) $ for all $ x,\ y\in N $ such that $ f(1)=3 $ and $ \sum\limits_{x=1}^{n}{f(x)=120} $ . Then the value of $ n $ is
[IIT 1992]
Options:
A) 4
B) 5
C) 6
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Since $ f(x+y)=f(x)f(y) $ for all $ x,\ y\in N $ , therefore for any $ x\in N $ $ f(x)=f(x-1+1)=f(x-1)f(1) $ $ =f(x-2){{[f(1)]}^{2}}=…….={{[f(1)]}^{x}} $
$ \Rightarrow $ $ f(x)=3^{x} $ , $ (\because \ f(1)=3) $ Now $ \sum\limits_{x=1}^{n}{f(x)=120} $
$ \Rightarrow $ $ \sum\limits_{x=1}^{n}{3^{x}=120} $
$ \Rightarrow $ $ \frac{3(3^{n}-1)}{(3-1)}=120 $
$ \Rightarrow $ $ n=4 $ .
BETA
