Given, equation is:
5f(x+y)=f(x)⋅f(y)...(1)
At x=0,y=0, we get
5f(0)=f(0)2⇒f(0)=5
Put y=1 in (1), then we get
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)
So,
f(1)=4f(0)=4⋅5f(2)=4f(1)=42⋅5f(3)=4f(2)=43⋅5f(4)=4f(3)=44⋅5f(5)=4f(4)=45⋅5
n=0∑5f(n)=5+5⋅4+5⋅42+5⋅43+5⋅44+5⋅45
⇒n=0∑5f(n)=35(46−1)=6825