Question

If \(\mathrm{P}\) and \(\mathrm{Q}\) are coefficients of \(\mathrm{x}^{\mathrm{n}}\) in expansion of \((1+\mathrm{x})^{2 \mathrm{n}}\) and \((1+x)^{2 n-1}\) then (a) \(\mathrm{P}=\mathrm{Q}\) (b) \(P=2 Q\) (c) \(2 \mathrm{P}=\mathrm{Q}\) (a) \(P+Q=0\)

Short answer

The relationship between P and Q is P = 2Q, which corresponds to option (b).

Step-by-step solution

Identify the coefficients of x^n in each expansion

In the given expansions, we have (1 + x)^(2n) and (1 + x)^(2n-1). We're looking for the coefficients of x^n in each expansion. Using the binomial theorem, we can write: For (1 + x)^(2n): P = (2n)Cn * 1^(2n-n) * x^n For (1 + x)^(2n-1): Q = (2n-1)Cn * 1^(2n-1-n) * x^n Now we'll determine the relationship between P and Q.

Express P and Q using binomial coefficients

We know that P = (2n)Cn and Q = (2n-1)Cn. We can write this as: P = \( \dfrac{(2n)!}{n!n!} \) Q = \( \dfrac{(2n-1)!}{n!(n-1)!} \)

Determine the relationship between P and Q

To find the relationship between P and Q, divide P by Q and simplify: \( \dfrac{P}{Q} = \dfrac{\frac{(2n)!}{n!n!}}{\frac{(2n-1)!}{n!(n-1)!}} \) Simplify this expression by cancelling out common factors: \( \dfrac{P}{Q} = \dfrac{(2n)! (n-1)!}{(2n-1)! n!} \) Now, use the property that n! = n(n-1)! to further simplify the expression: \( \dfrac{P}{Q} = \dfrac{2n(2n-1)! (n-1)!}{(2n-1)! n!} \) The (2n-1)! terms cancel out: \( \dfrac{P}{Q} = \dfrac{2n(n-1)!}{n!} \) Finally, use the same property n! = n(n-1)! to simplify: \( \dfrac{P}{Q} = \dfrac{2n}{n} = 2 \) So, P = 2Q. The correct answer is option (b).

Key concepts

Binomial Coefficients

Binomial coefficients play a critical role in combinatorics, especially when expanding binomial expressions using the Binomial Theorem. In essence, the binomial coefficient for a term in the expansion of \((1 + x)^n\) is calculated using the formula: \[{n \choose k} = \frac{n!}{k!(n-k)!}\]These coefficients are the numbers that appear in Pascal's triangle, each representing a count of combinations. In particular, these coefficients determine two critical quantities: how many ways we can choose \(k\) elements out of \(n\), and the weight each term in the binomial expansion carries.

• In expressions like \((1+x)^{2n}\), binomial coefficients \({2n \choose n}\) are used to find the coefficient of \(x^n\).
• These coefficients help us understand patterns and structures within algebraic expressions.

Understanding them is fundamental to solving various algebraic and combinatorial problems efficiently.
They help break down complex expansions into manageable components.

Combinatorics

Combinatorics is a branch of mathematics focused on counting, arrangement, and combination of objects. It forms a foundation for understanding binomial coefficients and is crucial in determining the number of ways elements can be selected or arranged. In the given problem, we use combinatorics to calculate the binomial coefficient for terms like \(x^n\) in binomial expansions. By knowing how to manipulate combinations, expressed as \({n \choose r}\), you can easily extract these coefficients from binomial expressions like \((1 + x)^{2n}\) and \((1 + x)^{2n-1}\).

• In our context, combinatorics allows us to find \(P = \frac{(2n)!}{n!n!}\) and \(Q = \frac{(2n-1)!}{n!(n-1)!}\).
• These calculations essentially count the number of ways to arrange or select elements, linking back to binomial coefficients.

Combinatorial techniques are not only used in theoretical computations but also in areas such as probability and data organization.

Algebraic Expansions

Algebraic expansions involve expanding expressions into a sum of terms. The Binomial Theorem provides a systematic way to expand binomial expressions raised to any positive integer power. The theorem states that:\[ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k \]Understanding this allows us to decompose expressions like \((1 + x)^{2n}\) into individual terms with coefficients determined by combinations.
For example, to find the coefficient \(P\) from \((1+x)^{2n}\), use the formula associated with binomial expansions to ascertain which term contains \(x^n\) and its coefficient. Each term in the series is influenced by both the binomial coefficient and any coefficient from the variables themselves. This method not only aids in identifying specific coefficients but also in constructing a complete polynomial from its root form.
The practical application of this process includes problem-solving in calculus and the simplification of polynomial expressions.