Problem 30 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(P_{n}\) denotes the product of the bipomial coefficients in the expansion of \((1+x)^{n}\), then \(\frac{P_{n+1}}{P_{n}}\) equals \(\begin{array}{ll}\text { (A) } \frac{(n+1)^{n}}{n !} & \text { (B) } \frac{n^{n}}{n !}\end{array}\) (C) \(\frac{(n+1)^{n}}{(n+1) !}\) (C) \(\frac{(n+1)^{n+1}}{(n+1) !}\)

Step-by-Step Solution

Answer

(C) \( \frac{(n+1)^{n}}{(n+1)!} \)

Step 1: Understand the Binomial Coefficients Product

The binomial expansion of \((1+x)^n\) is given by \( \sum_{k=0}^{n} \binom{n}{k}x^k \). The product of the binomial coefficients for \((1+x)^n\), denoted by \(P_n\), is \( \prod_{k=0}^{n} \binom{n}{k} \).

Step 2: Calculate Product of Binomial Coefficients

The expression for \(P_n\) is the product of all coefficients in the expansion. Using the identity \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), we get: \[ P_n = \prod_{k=0}^{n} \frac{n!}{k!(n-k)!} = \frac{(n!)^{n+1}}{(0!)^2(1!)^2...(n!)^2} \]

Step 3: Simplify \(P_{n+1}\) Compared to \(P_n\)

The product \(P_{n+1}\) for the expansion \((1+x)^{n+1}\) is \( \frac{((n+1)!)^{n+2}}{(0!)^2(1!)^2...(n+1)!^2} \).

Step 4: Evaluate \(\frac{P_{n+1}}{P_n}\)

The ratio \( \frac{P_{n+1}}{P_n} \) simplifies to \[ \frac{((n+1)!)^{n+2}/((0!)^2(1!)^2...(n+1)!^2)}{(n!)^{n+1}/((0!)^2(1!)^2...n!^2)} = \frac{(n+1)^{n}}{(n+1)!} \].

Step 5: Conclusion

The calculation shows that \( \frac{P_{n+1}}{P_n} = \frac{(n+1)^{n}}{(n+1)!} \). Hence, the correct answer is (C).

Key Concepts

Binomial Coefficients

In the world of mathematics, binomial coefficients play a crucial role in the expansion of binomial expressions. A binomial expression typically looks like \((1+x)^n\). When expanded using the binomial theorem, it results in a series: \( \sum_{k=0}^{n} \binom{n}{k} x^k \). Here, \( \binom{n}{k} \) is a binomial coefficient, defined as the number of ways to choose \(k\) elements from a set of \(n\) elements without regard to order. These coefficients are conveniently represented using factorials and known as combinations.

Binomial coefficients are calculated using the formula:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

The exclamation mark \(!\) denotes a factorial, which is the product of all positive integers up to a given number. For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Understanding binomial coefficients is key to deciphering binomial expansions which appear frequently in statistical models and algebraic calculations.

Product of Coefficients

Calculating the product of binomial coefficients involves multiplying all the coefficients from the expansion of a binomial. To understand this, consider the expansion: \((1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k\). Here, \(P_n\) denotes the product of the coefficients:\

• \(P_n = \prod_{k=0}^{n} \binom{n}{k} \)

Using the identity \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), this product becomes more elaborate as \( \frac{(n!)^{n+1}}{(0!)^2(1!)^2\ldots(n!)^2}\). This expression looks complicated, but it's simply the result of multiplicatively combining each coefficient from the expanded form. By taking the factorial of \(n\) multiple times and dividing by all smaller factorials squared, each coefficient is reflected in the total product \(P_n\), showing a deeper look into the symmetry and structure within binomial expansions.

Mathematical Expansion

Mathematical expansion is the process of entering a general form into a specific sequence or set of terms. Expanding a binomial expression like \((1+x)^n\) leverages the binomial theorem, giving a sequence of terms where each term involves a binomial coefficient. This is important in multiple mathematical and practical fields because it provides a structured way to break down complex equations.

The purpose of expanding a binomial is often to simplify the problem by transforming it into a sum of calculated terms. In our exercise, by comparing \(P_{n+1}\) with \(P_n\), students solve \(\frac{P_{n+1}}{P_n}\), identifying optimal ways to manipulate and simplify expressions. The result \(\frac{(n+1)^{n}}{(n+1)!}\) shows how to reach a conclusion by methodically expanding every step, like peeling away layers of complexity in a problem. Therefore, binomial expansion not only aids in algebraic simplification but also in calculating probabilities and predicting outcomes in statistical combinatorics.