Question

Suppose $f: R \rightarrow(0, \infty)$ be a differentiable function such that $5 f(x+y)=f(x) \cdot f(y), \forall x, y \in R$ If $f(3)=320$, then $\displaystyle\sum_{n=0}^5 f( n )$ is equal to :

• A. 6575
• B. 6825
• C. 6875
• D. 6525

Answer 1

Answer: (B)

6825

Answer 2

5f(x+y)=f(x)⋅f(y)
5f(0)=f(0)2⇒f(0)=5
5f(x+1)=f(x)⋅f(1)
⇒f(x)f(x+1)​=5f(1)​
⇒f(0)f(1)​⋅f(1)f(2)​⋅f(2)f(3)​=(5f(1)​)3
⇒5320​=53(f(1))3​⇒f(1)=20
∴5f(x+1)=20⋅f(x)⇒f(x+1)=4f(x)
n=0∑5​f(n)=5+5.4+5.42+5.43+5.44+5.45
=35[46−1]​=6825