Problem 164 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(A=1+r^{a}+r^{2 a}+\ldots \ldots \ldots\) to \(\infty\) and \(B=1+r^{b}+r^{2 b}+\ldots \ldots \ldots\) to \(\infty\), prove that \(r=\left(\frac{A-1}{A}\right)^{1 / a}=\left(\frac{B-1}{B}\right)^{1 / b}\)

Step-by-Step Solution

Answer

After applying the sum formula of an infinite geometric series to \( A \) and \( B \), we can rearrange to solve for \( r \) which gives us \( r = \left( \frac{A-1}{A} \right) ^ {1/a} = \left(\frac{B-1}{B}\right) ^ {1/b} \).

Step 1: Identify the Infinite Geometric Series

First, note that both \( A \) and \( B \) represent the sum of infinite geometric series, with \( r^a \) and \( r^b \) as the common ratios, respectively.

Step 2: Apply Geometric Series Sum Formula

The sum of an infinite geometric series with initial term \( a \) and common ratio \( r \) is given by \( S = a/(1-r) \). Therefore, we can write \( A = 1/(1-r^a) \) and \( B = 1/(1-r^b) \).

Step 3: Solve Equations for r

Now, solve the equations for \( r \). This gives \( r^a = (A - 1)/A \) and \( r^b = (B - 1)/B \).

Step 4: Take Reciprocal

Taking the reciprocal of these equations, we get \( r = (\frac{A-1}{A}) ^ {1/a} \) and \( r = (\frac{B-1}{B}) ^ {1/b} \).

Key Concepts

Geometric Series

A geometric series is a fascinating mathematical concept where each term is derived by multiplying the previous term by a fixed non-zero number, known as the common ratio. Consider a basic example where you have a starting number, and you multiply it by a specific factor repeatedly.
For instance, the series 1, 2, 4, 8, ... is geometric because each term is twice the previous term, giving it a common ratio of 2.

In the context of the exercise, we explore an infinite geometric series which expands indefinitely. Here, the series written as \(1 + r^a + r^{2a} + \ldots \) shows how each subsequent term is generated by raising the common ratio \(r\) to higher powers, progressively.
This property makes geometric series both intriguing and inherently convergent under certain conditions, especially when dealing with infinite terms.

Sum Formula

The sum formula for a geometric series allows us to calculate the total sum of all the terms in the series. For a finite geometric series, this is relatively straightforward, but it becomes especially interesting when applied to infinite series.

For an infinite geometric series beginning with a term \(a\) and having a common ratio \(r\) where \(|r| < 1\), the sum \(S\) is derived using the formula: \[ S = \frac{a}{1-r} \]
This formula is pivotal in solving problems involving infinite geometric sums, as it shows that the series converges to a fixed value.

• In our exercise, applying this formula results in expressions like \( A = \frac{1}{1 - r^a} \) and \( B = \frac{1}{1 - r^b} \), where \(A\) and \(B\) symbolize the sums of their respective infinite series.

This simple yet powerful tool allows us to manipulate series sums, making it easier to draw conclusions in mathematical proofs.

Infinite Series

An infinite series is a sum of infinitely many terms. The concept seems overwhelming at first—how can you add an infinite number of values? Yet, mathematics provides us with tools to deal with such challenges.
When considering infinite geometric series specifically, convergence is often a key focus. If the terms of the series shrink ever smaller, you can actually sum them to a finite limit.

In the context of this exercise, both series \(A\) and \(B\) were infinite geometric series. The statements \( A = \frac{1}{1 - r^a} \) and \( B = \frac{1}{1 - r^b} \) demonstrate how the sums of infinite terms can be represented with finite results under convergence conditions.

• The ability to sum to infinity may seem like magic, but it's mathematically possible when the series terms decrease enough, as highlighted by \(|r| < 1\) condition.
• This is why taking the limit becomes essential, leading us to the conclusion for \(r\), as shown in the final result \( r = (\frac{A-1}{A})^{1/a} = (\frac{B-1}{B})^{1/b} \).

Understanding how infinite series work is crucial for grasping complex mathematical ideas across various fields.