Problem 6 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The two consecutive terms in the expansion of \((3 x+2)^{74}\), whose coefficients are equal, are (A) 20 th and 21 st (B) 30 th and 31 st (C) 40 th and 41 st (D) none of these

Step-by-Step Solution

Answer

The consecutive terms with equal coefficients are the 30th and 31st terms (Option B).

Step 1: Understanding the Problem

We need to find two consecutive terms in the expansion of \((3x + 2)^{74}\) that have equal coefficients. The expression is a binomial expansion using the Binomial Theorem.

Step 2: Apply the Binomial Theorem

According to the Binomial Theorem, the general term \(T_{r+1}\) in the expansion of \((3x + 2)^{74}\) is given by: \[T_{r+1} = \binom{74}{r} (3x)^{74-r} \cdot 2^{r}\].Our task is to find when the coefficients of two consecutive terms \(T_{r+1}\) and \(T_{r+2}\) are equal.

Step 3: Set Up the Equality of Coefficients

The coefficients of the consecutive terms are equal if:\[\binom{74}{r} \cdot 3^{74-r} \cdot 2^r = \binom{74}{r+1} \cdot 3^{74-(r+1)} \cdot 2^{r+1}.\]This simplifies to:\[\binom{74}{r} \cdot 3^{74-r} \cdot 2^r = \binom{74}{r+1} \cdot 3^{73-r} \cdot 2^{r+1}.\]

Step 4: Simplify the Equation

Divide both sides by common terms:\[\frac{\binom{74}{r}}{\binom{74}{r+1}} \cdot 3 = 2.\]Further simplify using the properties of binomials:\[\frac{(r+1)}{75-r} \cdot 3 = 2.\]

Step 5: Solve for r

Solve the equation \[(r+1) \cdot 3 = 2(75-r)\] to find \(r\):\[3r + 3 = 150 - 2r,\]\[5r = 147,\]\[r = 29.4.\]Since \(r\) must be an integer, round \(r\) to the nearest integer which is 30.

Step 6: Verify the Solution

Since \(r\) begins from 0, the 31st term (\(r+1=31\)) and the 30th term (\(r=30\)) are indeed consecutive terms. Check options to see if 30th and 31st positions match an option. Note that our estimate matches with Option B.

Key Concepts

Binomial Expansion

The binomial expansion is a crucial concept in mathematics, particularly in algebra, that explains how to expand expressions raised to a power, such as \( (a + b)^n \). The Binomial Theorem provides a method to perform this expansion. It states that: \[(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}.\] Here, \(\binom{n}{r}\) is a binomial coefficient, calculated as \(\frac{n!}{r!(n-r)!}\), which tells us how many different ways we can pick \(r\) elements from a total of \(n\). For each term in the series, \(a^{n-r}\) and \((b^r)\) follow based on each choice of \(r\). Understanding this allows us to find any specific term in the expansion without having to expand the whole expression, making computations more manageable, especially for high-degree polynomials. In our original problem, we applied this concept to expand \( (3x+2)^{74}\) to find the specific coefficients of terms.

Consecutive Terms

In a binomial expansion, consecutive terms follow one after the other seamlessly. These are represented typically as \(T_{r+1}\) for the term based on \(\binom{n}{r}\), followed by \(T_{r+2}\) based on \(\binom{n}{r+1}\). Understanding the position and relationship between these terms is crucial in problems where comparisons are required, like ensuring the equality of coefficients. For consecutive terms in expansions, the difference between their corresponding \(r\) values is exactly one. This implies that their terms are structurally very similar, often differing by a factor coming from the coefficients or the variable powers.In our given task, identifying two consecutive terms with equal coefficients involves setting their coefficients equal to each other and working through the algebra to find the suitable position \(r\). This is how we narrowed down the 30th and 31st terms in the given expansion example.

Coefficient Equality

The notion of coefficient equality appears frequently in binomial expansion problems. It describes a situation where the coefficients of two consecutive terms in the expansion are identical. To find this, an equation is set up equating the coefficients derived from the binomial terms.For our exercise concerning \( (3x+2)^{74}\), it involved determining the conditions under which the coefficients \(\binom{74}{r} \cdot 3^{74-r} \cdot 2^r\) and \(\binom{74}{r+1} \cdot 3^{73-r} \cdot 2^{r+1}\) would be equal.Through simplification, by cancelling out common elements and simplifying the ratios, we derived \(\frac{(r+1)}{75-r} \cdot 3 = 2\). Solving this equation helps us pinpoint the appropriate \(r\) value where the consecutive terms, specifically the 30th and 31st, have equal coefficients. This very practical application helps solve not just homework problems but also supports deeper insights into binomial functions.