Problem 22 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the coefficients of the three successive terms in the binomial expansion of \((1+x)^{n}\) are in the ratio \(1: 7: 42\), then the first of these terms in the expansion is: (a) \(8^{\text {th }}\) (b) \(6^{\text {th }}\) (c) \(7^{\text {? }}\) (d) \(9^{\text {th }}\)

Step-by-Step Solution

Answer

The first term is the \(7^{\text{th}}\) term.

Step 1: Understanding Binomial Expansion

The binomial expansion of \((1+x)^n\) is given by \(\sum_{r=0}^{n} \binom{n}{r} x^r\), where \(\binom{n}{r}\) is the binomial coefficient \(\frac{n!}{r!(n-r)!}\). For this problem, the focus is on the coefficients of the terms.

Step 2: Expressing the Ratios

The given ratio of the coefficients is \(1: 7: 42\). Let the terms associated with these coefficients be \(\binom{n}{r-1}\), \(\binom{n}{r}\), and \(\binom{n}{r+1}\). Hence, \(\frac{\binom{n}{r}}{\binom{n}{r-1}} = 7\) and \(\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{42}{7} = 6\).

Step 3: Setting Up Equations

Calculate the first equation:\[ \frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r} = 7 \Rightarrow n-r+1 = 7r \]Rearranging gives:\[ n = 8r - 1 \]Now calculate the second equation:\[ \frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1} = 6 \Rightarrow n-r = 6(r+1) \\Rightarrow n-r = 6r + 6 \]Rearranging gives:\[ n = 7r + 6 \]

Step 4: Solving the Equations

From the equations \(n = 8r - 1\) and \(n = 7r + 6\), equate the two expressions for \(n\):\[8r - 1 = 7r + 6 \]Solve for \(r\):\[8r - 7r = 6 + 1 \r = 7\]

Step 5: Conclusion of Steps

Thus, the first of the successive terms with coefficients in the ratio \(1: 7: 42\) is the \(7^{\text{th}}\) term in the expansion. This corresponds to option (c).

Key Concepts

Binomial Coefficient

In the binomial expansion of \((1+x)^n\), the binomial coefficient \(\binom{n}{r}\) quantifies the number of ways to choose \(r\) elements from a set of \(n\) elements. It is calculated using the formula \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\). Binomial coefficients are key in indicating the strength or weight of each term in a binomial expansion.
Here, each coefficient tells us how each term in the expansion contributes to the overall polynomial. For example, in the expansion \((1 + x)^n\), understanding which terms have large coefficients helps us grasp which terms are most influential for a particular value of \(x\). The binomial coefficient's value depends on both \(n\) and \(r\), as it describes the ratio of factorials, providing a combinatorial count.
It is this fascinating nature of combinatorial distribution that underpins the importance of choosing the right terms when interpreting polynomial expressions. The coefficients come from iteratively applying the binomial theorem which is foundational in algebra.

Successive Terms Ratio

When looking at successive terms in a binomial expansion, the ratio between terms tells us how the influence of each term evolves. Given a ratio of 1: 7: 42 for terms in a binomial expansion, we are asked to determine the position of these terms.
The ratio of successive terms is given by comparing the binomial coefficients of two neighboring terms. If we have coefficients \(\binom{n}{r-1}, \binom{n}{r}, \binom{n}{r+1}\), the ratio \(\frac{\binom{n}{r}}{\binom{n}{r-1}} = 7\) indicates that the \(r\)-th term is seven times larger than the \((r-1)\)-th term. Similarly, \(\frac{\binom{n}{r+1}}{\binom{n}{r}} = 6\) means the \((r+1)\)-th term is six times bigger than the \(r\)-th term.
These equations are fundamental for calculating the exact succession of terms, allowing us to trace changes and trends in the polynomial expansion. They are practical tools grounded in the principles of algebra and calculus.

Coefficient Comparison

Comparing coefficients is a crucial part of solving problems in binomial expansion. When we analyze ratios like 1: 7: 42 within coefficients, we're looking at how different terms compare and what it implies about their arrangement.
In the solution, equating ratios to their respective expressions, \(\frac{\binom{n}{r}}{\binom{n}{r-1}} = 7\), and \(\frac{\binom{n}{r+1}}{\binom{n}{r}} = 6\), allowed us to form two equations: \(n = 8r - 1\) and \(n = 7r + 6\). Solving these together gives the precise term positioning in respect to the sequence.
Equations are set up and solved to identify term positions like the 7th term in this exercise. This method of comparing coefficients is pivotal, aiding in finding which terms hold the first position among specified ratios. It's a clear, systematic approach enabled by algebraic manipulation.