Question

If f(1)=2 and f(n+1) = f(n)^2+4 then find the value of f(3).

Answer 1

Answer:

Given that, f(1)=1, f(2n)=f(n), f(2n+1)={f(n)}

2

−2

Now, f(2)=f(2×1)=f(1)=1,

f(3)=f(2×1+1)

={f(1)}

2

−2=1−2=−1,

f(4)=f(2×2)=f(2)=1

f(5)=f(2×2+1)

={f(2)}

2

−2=1−2=−1

and so on

Now, f(1)+f(2)+⋯+f(25)=1+1−1+1−1+⋯+(−1)=1