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If P_n denotes the product of all the coefficients in the expa

Question

If $P_{n}$ denotes the product of all the coefficients in the expansion of $(1 + x)^{n}$ , then $\frac{P _{n + 1}}{P _{n}}$ is equal to:

\frac{( n + 2 ) ^{n}}{n !}

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

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

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

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Isabella Discussed

P_{n}

denotes the product of all the coefficients in the expansion of

(1 + x)^{n}

, then

\frac{P _{n + 1}}{P _{n}}

is equal to:

9 mins ago

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9 mins ago

Text solutionVerified

(1 + x)^{n}

P_{n + 1} =_{n + 1} C_{0} \times_{n + 1} C_{1} \times_{n + 1} C_{2} \times ........ \times_{n + 1} C_{n + 1} P_{n} =_{n} C_{0} \times_{n} C_{1} \times_{n} C_{2} \times ........ \times_{n} C_{n}

\prod_{r = 0}^{n} \frac{_{n + 1} C _{r}}{_{n} C _{r}} (∵_{n + 1} C_{n + 1} = 1) = \prod_{r = 0}^{n} \frac{( n + 1 )!}{r ! ( n + 1 - r )!} \times \frac{r ! ( n - r )!}{n !} = \prod_{r = 0}^{n} \frac{( n + 1 )}{( n - r + 1 )} = (n + 1)^{n} \prod_{r = 0}^{n} \frac{1}{n - r + 1} = \frac{( n + 1 ) ^{n}}{( n + 1 ) !}

Hence, the correct option is

D

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Question Text	If $P_{n}$ denotes the product of all the coefficients in the expansion of $(1 + x)^{n}$ , then $\frac{P _{n + 1}}{P _{n}}$ is equal to:
Answer Type	Text solution:1
Upvotes	150