Question

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\limits_{x = 1}^n {f(x) = 120} $ find the value of $n?$
A. $4$
B. $2$
C. $1$
D. $3$

Answer 1

Hint: Here we have been given $f(x + y) = f(x)f(y)$ and $f(1) = 3$ . So first of all we will find the value of $f(2),f(3)$ and $f(4)$. After finding the above values, we will apply the formula given in the question. So we can write the expression $f(2)$ also as $f(1 + 1)$. It is given in the question that we can write $f(x + y) = f(x)f(y)$ . So by applying this we can also write $f(1 + 1)$ as $f(1)f(1)$ and it is given in the question that $f(1) = 3$. Therefore by putting the value and applying this we will solve the given question.

Complete step by step answer:
Here we have $f(x + y) = f(x)f(y)$ for all $x,y \in N$ such that $f(1) = 3$ and $\sum\limits_{x = 1}^n {f(x) = 120} $
We will calculate the value of $f(2)$ .
From the hint, we can see that we have $f(1)f(1)$. As given in the question that the value of $f(1) = 3$ , so we have:
$ \Rightarrow 3 \times 3 = {3^2}$
Similarly we can write $f(3)$ as $f(2 + 1)$. Again by applying the property we have $f(2)f(1)$ and by putting the value, it gives us,
$ \Rightarrow {3^2} \times 3 = {3^3}$
Now we will find the value of $f(4)$. By splitting the term it can be written as $f(3 + 1)$. Again we can write this function as the formula i.e. $f(3)f(1)$.

By substituting the values, we have
$ \Rightarrow {3^3} \times 3 = {3^4}$
We can see that whenever we try to calculate the value of a function, we get the same power on $3$ .
So we can write this also as
$ \Rightarrow f(n) = {3^n}$
Given in the question that
$ \Rightarrow \sum\limits_{x = 1}^n {f(x) = 120} $ .
We have to find the value of the sum of the series up to $n$ terms, so here we can say that $x = n$. By putting this we can also write this as
$ \Rightarrow \sum {f(n)} = 120$ .

We can break this sum and it can be written as:
$ \Rightarrow f(1) + f(2) + f(3) + ... + f(n) = 120$
By substituting the values from the above we can write:
$ \Rightarrow 3 + {3^2} + {3^3} + ... + {3^n} = 120$
We can see that the left side of the equation is in Geometric Progression, here we have $a = 3$ and $ \Rightarrow r = \dfrac{{{3^2}}}{3} = 3$
Now by applying the formula of sum of G.P, we have
$ \Rightarrow \dfrac{{3({3^n} - 1)}}{{3 - 1}} = 120$
On simplifying we can write
$\dfrac{{3({3^n} - 1)}}{2} = 120 \Rightarrow 3({3^n} - 1) = 2 \times 120$

By arranging the terms , we can write
$ \Rightarrow {3^n} - 1 = \dfrac{{240}}{3} \\
\Rightarrow {3^n} - 1 = 80$
We can write the expression as
$ \Rightarrow {3^n} = 80 + 1 = 81$
We can write $81$ as the product of three i.e.
$ \Rightarrow 3 \times 3 \times 3 \times 3 = {3^4}$
By putting this in the expression, we have
$ \Rightarrow {3^n} = {3^4}$
Since the bases on both sides are the same, we can eliminate them. It gives us
$\therefore n = 4$

Hence the correct option is A.

Note: We should note that the formula that we have used before in the solution is,
$ \Rightarrow {S_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}}$
We should know that G.P is a sequence where each new term is obtained by multiplying the preceding term by a constant $r$ , known as common ratio. The general form of G.P is,
$a,ar,a{r^2},a{r^3}...a{r^{n - 1}}$
And the sum of the terms of G.P is,
$ \Rightarrow {S_n} = a + ar + a{r^2} + a{r^3} + ...$