If Pn denotes the product of the binomial coefficients in the expansion of (1+x)n, then (Pn+1/Pn) equals

Question

If Pn denotes the product of the binomial coefficients in the expansion of (1+x)n, then (Pn+1/Pn) equals (A)  (n+1/n !)  (B)  (nn/n !)  (C)  ((n+1)n/(n+1) !)  (D)  ((n+1)n+1/(n+1) !)

Tardigrade Admin · Accepted Answer

Here, Pn = nC0. nC1. nC2 .... nCn and Pn+1 = n+1C0. n+1C1. n+1C2 .... n+1Cn+1 ∴ (Pn+1/Pn) = (n+1C0. n+1C1. n+1C2 .... n+1Cn+1/nC0. nC1. nC2 .... nCn) = ( (n+1C1/nC0))( (n+1C2/nC1))( (n+1C3/nC2)) ..... ((n+1Cn+1/nCn)) = ((n+1/1))((n+1/2))((n+1/3)).....((n+1/n+1)) = ((n+1)n+1/(n+1) !)

Q. If $P_{n}$ denotes the product of the binomial coefficients in the expansion of $(1 + x)^{n}$ , then $\frac{P _{n + 1}}{P _{n}}$ equals

$\frac{n + 1}{n !}$

$\frac{n ^{n}}{n !}$

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

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

Q. If Pn​ denotes the product of the binomial coefficients in the expansion of (1+x)n, then Pn​Pn+1​​ equals

n!n+1​

n!nn​

(n+1)!(n+1)n​

(n+1)!(n+1)n+1​

Q. If $P_{n}$ denotes the product of the binomial coefficients in the expansion of $(1 + x)^{n}$ , then $\frac{P _{n + 1}}{P _{n}}$ equals

$\frac{n + 1}{n !}$

$\frac{n ^{n}}{n !}$

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

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