Problem 25 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f\) be a function satisfying \(f(x+y)=f(x) f(y)\) for all \(x, y \in \mathbb{R}\). If \(f(1)=3\), then \(\sum f(r)\) is equal to (A) \(\frac{3}{2}\left(3^{n}-1\right)\) (B) \(\frac{3}{2} n(n+1)\) (C) \(3^{n+1}-3\) (D) None of these

Step-by-Step Solution

Answer

The sum \(\sum f(r)\) is \(\frac{3}{2}(3^n - 1)\), which is option (A).

Step 1: Understand the Functional Equation

We are given that the function \(f\) satisfies the equation \(f(x+y) = f(x) f(y)\) for all real numbers \(x\) and \(y\). This is a known exponential functional equation. Based on standard results, we consider \(f(x) = a^x\) where \(a\) is a constant.

Step 2: Use the Initial Condition

We were given that \(f(1) = 3\). Plugging \(1\) into our assumed function form \(f(x) = a^x\), we get \(a^1 = 3\). Thus, \(a = 3\). This implies \(f(x) = 3^x\).

Step 3: Express the Sum Using \(f(x)\)

Since \(f(x) = 3^x\), we need to compute \(\sum f(r)\). Assuming the sum is over a fixed set like natural numbers up to \(n\), this implies calculating \(\sum_{r=1}^n f(r) = \sum_{r=1}^n 3^r\).

Step 4: Calculate the Sum of a Geometric Series

The sum \(\sum_{r=1}^n 3^r\) is a geometric series with the first term \(a = 3\) and a common ratio \(r = 3\), for \(n\) terms. The formula for the sum of a geometric series is given by \(S_n = a \frac{r^n - 1}{r - 1}\). Substituting \(a = 3\) and \(r = 3\), we have \(S_n = 3 \frac{3^n - 1}{3 - 1}\).

Step 5: Simplify the Series Sum Expression

Simplifying the expression for \(S_n\), we get \(S_n = 3 \frac{3^n - 1}{2} = \frac{3}{2}(3^n - 1)\). This matches option (A).

Key Concepts

Exponential Function

In the context of functional equations, an exponential function is a type of mathematical function characterized by its dependence on an exponent. It is crucial to determine that a function like \( f(x) = a^x \) will fit an exponential functional equation such as \( f(x+y) = f(x)f(y) \). This kind of equation implies a very specific growth pattern, often related to powers of a base number like \( a \). The base \( a \) determines how quickly or slowly the function grows. In our problem, we found out that \( f(1) = 3 \), meaning \( a = 3 \) and thereby \( f(x) = 3^x \). This understanding is key to utilizing our function in calculating sums or series that follow this growth rule.

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. When solving our problem, we examined the series resulting from \( f(x) = 3^x \), where each term is 3 times the one before it. This makes it a perfect example of a geometric series.

• The first term in our series is \( a = 3 \).
• The common ratio \( r \) is also \( 3 \).

This type of series is significant because we can sum it efficiently using a standard formula. Recognizing a geometric series helps us apply specific techniques to find the sum.

Sum of Series

The sum of a series is an essential concept when working with functions satisfying exponential functional equations. It helps determine the total value when adding up all the terms in a series. The series \( \sum_{r=1}^n 3^r \) is summed using the formula for a finite geometric series:

• \( S_n = a \frac{r^n - 1}{r - 1} \)
• Our series has \( a = 3 \) and \( r = 3 \).

Substituting these values, the sum becomes \( S_n = 3 \frac{3^n - 1}{2} = \frac{3}{2}(3^n - 1) \). Breaking down the sum into smaller steps is crucial for understanding and reinforcing the core concept. Thus, Option (A) is the correct expression, reflecting the sum of our series correctly and offering insight into applications of geometric series in solving such equations.