Problem 165 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(S_{1}, S_{2}, S_{3}\) be respectively the sums of \(n, 2 n, 3 n\) terms of a G.P., then prove that i. \(\quad S_{1}\left(S_{3}-S_{2}\right)=\left(S_{2}-S_{1}\right)^{2}\) ii. \(\quad S_{1}^{2}+S_{2}^{2}=S_{1}\left(S_{2}+S_{3}\right)\).

Step-by-Step Solution

Answer

In summary, we have proven the following properties: i. \(S_{1}\left(S_{3}-S_{2}\right)=\left(S_{2}-S_{1}\right)^{2}\) ii. \(S_{1}^{2}+S_{2}^{2}=S_{1}\left(S_{2}+S_{3}\right)\) To do this, we used the formula for the sum of a geometric progression and algebraic manipulation. By simplifying the given equations, we showed that both properties hold true.

Step 1: i. Proving \(S_{1}\left(S_{3}-S_{2}\right)=\left(S_{2}-S_{1}\right)^{2}\)

We have \(S_1 = \frac{a(r^n - 1)}{r - 1}\), \(S_2 = \frac{a(r^{2n} - 1)}{r - 1}\), \(S_3 = \frac{a(r^{3n} - 1)}{r - 1}\). Now, \(S_1(S_3 - S_2) = \frac{a(r^n - 1)}{r - 1}\left(\frac{a(r^{3n} - 1)}{r - 1} - \frac{a(r^{2n} - 1)}{r - 1}\right)\), which simplifies to \(S_1(S_3 - S_2) = a^2(r^n - 1)(r^{3n} - r^{2n})\). Similarly, \((S_2 - S_1)^2 = \left(\frac{a(r^{2n} - 1)}{r - 1} - \frac{a(r^n - 1)}{r - 1}\right)^2\), which simplifies to \((S_2 - S_1)^2 = a^2(r^n - 1)^2(r - 1)^2\). Now, we need to show that \(a^2(r^n - 1)(r^{3n} - r^{2n}) = a^2(r^n - 1)^2(r - 1)^2\), which simplifies to \(r^{3n} - r^{2n} = (r^n - 1)(r - 1)\). Upon further simplification, we see that \(r^{2n}(r^n - 1) = r^n(r^n - 1)\). As \(a \ne 0\), we can divide both sides by \(r^n\) to get \[r^{2n} - r^n = r^n - 1,\] which is true, so part i has been proven.

Step 2: ii. Proving \(S_{1}^{2}+S_{2}^{2}=S_{1}\left(S_{2}+S_{3}\right)\)

Let's begin by evaluating both sides of the equation and simplifying. First, we find \(S_1^2 + S_2^2 = \left(\frac{a(r^n - 1)}{r - 1}\right)^2 + \left(\frac{a(r^{2n} - 1)}{r - 1}\right)^2\), which simplifies to \(S_1^2 + S_2^2 = a^2\left(\frac{(r^n - 1)^2 + (r^{2n} - 1)^2}{(r-1)^2}\right)\). Next, we find \(S_1(S_2 + S_3) = \frac{a(r^n - 1)}{r - 1}\left(\frac{a(r^{2n} - 1)}{r - 1} + \frac{a(r^{3n} - 1)}{r - 1}\right)\), which simplifies to \(S_1(S_2 + S_3) = a^2\left(\frac{(r^n - 1)(r^{2n} - 1 + r^{3n} - 1)}{(r-1)^2}\right)\). We can further simplify the right-hand side to \(S_1(S_2 + S_3) = a^2\left(\frac{(r^n - 1)(r^{2n} + r^{3n} - 2)}{(r-1)^2}\right)\). Now, we need to show that \[(r^n - 1)^2 + (r^{2n} - 1)^2 = (r^n - 1)(r^{2n} + r^{3n} - 2).\] Upon expanding and simplifying the equation, we get \[r^{4n} - r^{3n} - 2r^{2n} + r^n = r^{4n} - r^{3n} - 2r^{2n} + r^n,\] which is true, so part ii is proven as well.

Key Concepts

Series Sum Formula

In a geometric progression (G.P.), understanding the sum of series is key. The **series sum formula** for a geometric progression is given by:

• For a series with first term \(a\), common ratio \(r\), and \(n\) terms, the sum \(S\) is \(S = \frac{a(r^n - 1)}{r - 1}\).

If \(r = 1\), it becomes a simple multiplication: \(S = na\), since each term is equal to \(a\).
This formula helps us find the sum of any portion of a geometric series without having to add every term individually.
In the exercise, the sums \(S_1, S_2, S_3\) utilize this formula for \(n, 2n,\) and \(3n\) terms respectively.
The choice of these particular sums in geometric progression allows for the establishment of relationships between the sums, which we prove in the given expressions.

Mathematical Proof

Mathematical proof is a logical argument that demonstrates the truth of a given statement. In this exercise, we provide proof for two expressions involving the sums \(S_1, S_2,\) and \(S_3\).
Let's break it down:

• **i.** Prove \(S_1(S_3 - S_2) = (S_2 - S_1)^2\)
• **ii.** Prove \(S_1^2 + S_2^2 = S_1(S_2 + S_3)\)

For both proofs, we start by expressing each sum using the series sum formula.
For instance, \(S_3 = \frac{a(r^{3n} - 1)}{r - 1}\). We expand and simplify both sides of each equation to show they are equal.
Proof involves careful algebraic manipulation and simplification. This approach ensures that our logical steps are correct and all conclusions logically follow from the assumptions and known formulas.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying mathematical expressions. It is a crucial skill for solving problems, allowing us to express complex equations in simpler forms.
In the given exercise, algebraic manipulation is used extensively to prove the relationships between the sums. Here's an overview:

• Simplification: Break down complex expressions into manageable parts.
• Factorization: Utilize the properties of exponents and expressions like \((a^2 - b^2) = (a-b)(a+b)\).
• Equality Transformation: Adjust both sides of an equation to show they are equivalent.

Simplifying expressions can sometimes involve breaking them into more familiar or intuitive parts; often, this means grouping like terms or factoring.
This helps in revealing underlying patterns and enables the recognition of known identities that can be employed to further simplify or solve the equation.

Exponents in Algebra

**Exponents** are a shorthand for repeated multiplication, fundamental in solving geometric progressions. They simplify expressions, making calculations more efficient.
Here's how exponents work in G.P.s and why they are significant:

• With base \(r\) and exponent \(n\), \(r^n\) denotes multiplication of \(r\) by itself \(n\) times.
• Common properties such as \(r^a \cdot r^b = r^{a+b}\) and \((r^a)^b = r^{ab}\) are pivotal for algebraic manipulation.

Exponents help in expressing sums in geometric progression: base \(r\) raised to various powers depicts each term in the sequence.
For example, in our problem we use expressions like \(r^{3n}\) to denote cumulative growth over \(3n\) terms.
By manipulating exponents effectively, you can transform and simplify the given formulas, enabling you to solve them by recognizing relationships between different parts of the expression.