Question

If f(x) is a function satisfying f(x + y) = f(x)f(y) for all x, y ∈ N such that f(1) = 3 and $\sum _{x=1}^n$f(x) = 120. Then find the value of n.

Answer 1

Since f(x + y) = f(x) f(y) for all x, y ∈ N, therefore for any x ∈ N

f(x) = f(x - 1 + 1) = f(x - 1) f(1)

f(x - 2) ${\left[f\right(1\left)\right]}^2$ = ......... = ${\left[f\right(1\left)\right]}^x$

⇒ f(x) = $3^x$, ( ∵ f(1) = 3)

Now $\sum _{x=1}^n$f(x) = 120 ⇒ $\sum _{x=1}^n$ $3^x$ = 120

⇒ $\frac{3(3^n-1)}{3-1}$ = 120 ⇒ n = 4.