If (1 + x)^n = sumlimits_{r = 0}^n {{C_r}{x^r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the expression $P = \left( {1 + \frac{{{C_1}}}{{{C_0}}}} \right)\left( {1 + \frac{{{C_2}}}{{{C_1}}}} \right)....\left( {1 + \frac{{{C_n}}}{{

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the expression $P = \left( {1 + \frac{{{C_1}}}{{{C_0}}}} ight)\left( {1 + \frac{{{C_2}}}{{{C_1}}}} ight)....\left( {1 + \frac{{{C_n}}}{{{C_{n - 1}}}}} ight)$.We know that $\frac{{{C_r}}}{{{C_{r - 1}}}} = \frac{{n - r + 1}}{r}$.Substituting this into each term: $1 + \frac{{{C_r}}}{{{C_{r - 1}}}} = 1 + \frac{{n - r + 1}}{r} = \frac{{r + n - r + 1}}{r} = \frac{{n + 1}}{r}$.Thus,the product becomes:$P = \left( \frac{n + 1}{1} ight) \left( \frac{n + 1}{2} ight) \left( \frac{n + 1}{3} ight) .... \left( \frac{n + 1}{n} ight)$.$P = \frac{{{{(n + 1)}^n}}}{{1 	imes 2 	imes 3 	imes .... 	imes n}} = \frac{{{{(n + 1)}^n}}}{{n!}}$.

If $(1 + x)^n = \sum\limits_{r = 0}^n {{C_r}{x^r}} $,then $\left( {1 + \frac{{{C_1}}}{{{C_0}}}} \right)\left( {1 + \frac{{{C_2}}}{{{C_1}}}} \right)....\left( {1 + \frac{{{C_n}}}{{{C_{n - 1}}}}} \right) = $

Similar Questions

In the expansion of $(1 + x)^{50}$,the sum of the coefficients of odd powers of $x$ is

If $\sum\limits_{K = 1}^{12} {12K \cdot {^{12}C_K} \cdot {^{11}C_{K - 1}}} $ is equal to $\frac{{12 \times 21 \times 19 \times 17 \times \dots \times 3}}{{11!}} \times {2^{12}} \times p$,then $p$ is:

$3 \cdot C_0 + 7 \cdot C_1 + 11 \cdot C_2 + \ldots + (3 + 4n) C_n =$

If ${}^{20}C_{r}$ is the coefficient of $x^{r}$ in the expansion of $(1+x)^{20}$,then the value of $\sum_{r=0}^{20} r^{2} \cdot {}^{20}C_{r}$ is equal to:

If $\sum_{k=1}^{10} k^{2} \binom{10}{k}^{2} = 22000 L$,then $L$ is equal to $.....$

Vedclass Test Series

Exam Paper Generator

Online Exam Module