If f is a function satisfying f(x + y) = f(x) f(y) for all x, y ∈ N such that find the value of n.

Given that,

f (x + y) = f (x) × f (y) for all x, y ∈ N … (1)

f (1) = 3

Taking x = y = 1 in (1), we have

f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9

Similarly,

f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27

And, f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81

Thus, f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with the first term and common ratio both equal to 3.

We know that sum of terms in G.P is given by,

\( s_n=\dfrac{a(1-r^n)}{r-1}\)

And it’s given that,

\( \displaystyle\sum_{x=1}^{n}f(x)=120\)

Hence, the sum of terms of the function is 120.

\(120=\dfrac{3(3^n)-1}{3-1}\)

\(120=\dfrac{3}{2}(3^n-1)\)

3ⁿ - 1=80

3ⁿ = 81 = 3⁴

n = 4

Therefore, the value of n is 4.