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

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

Step by step solution

01

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.
02

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 = \( \dfrac{(2n-1)!}{n!(n-1)!} \)
03

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).

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