Question

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

Answer 1

Hint: We put $ y=1,2,3,4... $ in the given functional equation $ f\left( x+y \right)=f\left( x \right)f\left( y \right) $ and we get a geometric progression (GP ) whose sum of the terms is equal to the given summation $ \sum\limits_{x=1}^{n}{f\left( x \right)}=120 $ . We use the formula for sum of first $ n $ terms in GP as $ \dfrac{a\left( {{r}^{n}}-1 \right)}{r-1} $ and equate it to 120 to find $ n $ . \[\]

Complete step by step answer:
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 $ \left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},... $ is an GP, then
\[\dfrac{{{x}_{2}}}{{{x}_{1}}}=\dfrac{{{x}_{3}}}{{{x}_{1}}}...=r...(1)\]
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,
\[S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}\]

We are given in the question the summation value $ \sum\limits_{x=1}^{n}{f\left( x \right)}=120 $ which we can write by expansion as;
\[f\left( 1 \right)+f\left( 2 \right)+...+f\left( n \right)=120......\left( 1 \right)\]
We are also given the following functional equation.
\[f\left( x+y \right)=f\left( x \right)f\left( y \right).........\left( 2 \right)\]
We are also given $ f\left( 1 \right)=3 $ . Let us put $ y=1 $ in the above equation to have;
\[\begin{align}
  & f\left( 1+1 \right)=f\left( 1 \right)f\left( 1 \right) \\
 & \Rightarrow f\left( 2 \right)=3\cdot 3={{3}^{2}} \\
\end{align}\]
We put $ x=1,y=2 $ in equation (2) and use previously obtained $ f\left( 2 \right)={{3}^{2}} $ to have
\[\begin{align}
  & f\left( 2+1 \right)=f\left( 1 \right)f\left( 2 \right) \\
 & \Rightarrow f\left( 3 \right)=3\cdot {{3}^{2}} \\
 & \Rightarrow f\left( 3 \right)={{3}^{3}} \\
\end{align}\]
We can go on putting $ y=3,4,...n+1 $ to have values of the function as
\[f\left( 3 \right)={{3}^{4}},f\left( 4 \right)={{3}^{5}},...,f\left( n \right)={{3}^{n}}\]
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=\dfrac{{{3}^{2}}}{3}=\dfrac{{{3}^{3}}}{{{3}^{2}}}...=3 $ . We use formula for sum of first $ n $ terms of a GP and have;
\[\begin{align}
  & \Rightarrow \dfrac{3\left( {{3}^{n-1}}-1 \right)}{3-1}=120 \\
 & \Rightarrow \dfrac{3}{2}\left( {{3}^{n-1}}-1 \right)=120 \\
 & \Rightarrow {{3}^{n-1}}=120\times \dfrac{2}{3}+1=81 \\
 & \Rightarrow {{3}^{n-1}}={{3}^{4}} \\
\end{align}\]
We equate the exponents of both sides of the above equations to have;
\[\begin{align}
  & \Rightarrow n-1=4 \\
 & \therefore n=5 \\
\end{align}\]

Note:
We note that the question presumes $ x,y\in \mathsf{\mathbb{N}} $ and hence we have taken only natural numbers. An equation involving only functions is called a functional equation. If $ f $ is continuous and monotonic the for two operations $ {{o}_{1}},{{o}_{2}} $ and then Cauchy’s function equation is given by $ f\left( x{{o}_{1}}y \right)=f\left( x \right){{o}_{2}}f\left( y \right) $ . If $ {{o}_{1}} $ is multiplication and $ {{o}_{2}} $ is addition then $ f $ is an exponential function. Here in the problem the solution is $ f\left( x \right)={{3}^{x}} $ . If $ {{o}_{1}} $ is addition and $ {{o}_{2}} $ is multiplication then $ f $ is logarithmic function.