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

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

1. A
5
2. B
4
3. C
3
4. D
2

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Solution:

Geometric sequence otherwise known as Geometric progression, abbreviated as GP is a type sequence where the ratio between any two consecutive numbers is constant. If (xn)=x1,x2,x3,...is an GP, then
$\frac{{x}_{2}}{{x}_{1}}=\frac{{x}_{3}}{{x}_{1}}\dots .=r\dots ..$...(1)
Here r is called common ratio and if we take the first term ${x}_{1}=a$ , then sum of GP of first n terms is given by,
S = $\frac{a\left({r}^{n}-1\right)}{r-1}$
We are given in the question the summation value $\sum _{x=1}^{n}f\left(x\right)=120$which we can write by expansion as;
f(1) + f(2)+...+f(n)=120......(1)
We are also given the following functional equation.
f(x+y) = f(x)f(y).........(2)
We are also given f(1)=3. Let us put y=1 in the above equation to have;
f(1+1) = f(1)f(1)
f(2) = 33 =
We put x=1,y=2  in equation (2) and use previously obtained f(2)= ${3}^{2}$ to have
f(2+1) = f(1)f(2)
f(3) = 3${3}^{2}$
f(3) = ${3}^{3}$
We can go on putting y=3,4,...n+1 to have values of the function as
f(3)= ${3}^{4}$,f(4)= ${3}^{5}$...,f(n)= ${3}^{n}$
We put these values in equation (1) to have;
3+${3}^{2}$+...+${3}^{n}$=120
We see that in the left hand side of the above step, we have have GP with first term say a=3 and common ratio r=$\frac{{3}^{2}}{3}$=$\frac{{3}^{3}}{{3}^{2}}\dots \dots \dots$=3. We use formula for sum of first n terms of a GP and have;
= 120

We equate the exponents of both sides of the above equations to have;
n1= 4
∴ n= 5

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