Functions Question 572
Question: If $ f:R\to R $ satisfies $ f(x+y)=f(x)+f(y) $ , for all $ x,\ y\in R $ and $ f(1)=7 $ , then $ \sum\limits_{r=1}^{n}{f(r)} $ is
[AIEEE 2003]
Options:
A) $ \frac{7n}{2} $
B) $ \frac{7(n+1)}{2} $
C) $ 7n(n+1) $
D) $ \frac{7n(n+1)}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x+y)=f(x)+f(y) $
Put $ x=1,,y=0 $
Þ $ f(1)=f(1)+f(0)=7 $
Put $ x=1,,y=1 $
Þ $ f(2)=2.f(1)=2.7 $
Similarly $ f(3)=3.7 $ and so on
$ \therefore \sum\limits_{r=1}^{n}{f(r)=7,(1+2+3+…..+n)} $ = $ \frac{7n(n+1)}{2} $ .
BETA
