If f x is function satisfying f x - y - fx fy for all x- y- -N such that f1 - 3 and - x-1 n fx - 120- Then the value of n is-

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

Find the value of n, so that $\frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}}$ is the geometric mean between a and b

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

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If f(x) is function satisfying f(x+y)=f(x)f(y) for all x,y, ∈N such that f(1)=3 and ∑nx=1f(x)=120. Then the value of n is.

The correct option is D 4Given f(x+y)=f(x)f(y)∴f(x)=axUsing f(1)=3⇒a=3Thus f(x)=3x⇒∑nx=1f(x)=3(3n−1)3−1=120⇒3n=81⇒n=4Hence, option A is correct.

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

The correct option is D $4$
Given $f (x + y) = f (x) f (y)$

$∴ f (x) = a^{x}$

Using $f (1) = 3 \Rightarrow a = 3$

Thus $f (x) = 3^{x}$

$\Rightarrow \sum_{x = 1}^{n} f (x) = \frac{3 (3^{n} - 1)}{3 - 1} = 120$

$\Rightarrow 3^{n} = 81 \Rightarrow n = 4$

Hence, option $A$ is correct.