The value of the sum left({ }^{n} C_{1}right) | vedclass

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

Question

(D) We know that the binomial expansion is given by: $(1+x)^{n} = { }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\ldots+{ }^{n} C_{n} x^{n}$ and $(

Vedclass · Accepted Answer

${ }^{2 n} C_{n}-1$

Vedclass · Answer

(D) We know that the binomial expansion is given by: $(1+x)^{n} = { }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\ldots+{ }^{n} C_{n} x^{n}$ and $(x+1)^{n} = { }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1}+{ }^{n} C_{2} x^{n-2}+\ldots+{ }^{n} C_{n}$. Multiplying these two expressions, we get $(1+x)^{2n} = (1+x)^{n}(x+1)^{n}$. The coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is ${ }^{2n} C_{n}$. By multiplying the two series, the coefficient of $x^{n}$ is the sum of the products of corresponding coefficients: $({ }^{n} C_{0})^{2} + ({ }^{n} C_{1})^{2} + ({ }^{n} C_{2})^{2} + \ldots + ({ }^{n} C_{n})^{2} = { }^{2n} C_{n}$. The given sum starts from $({ }^{n} C_{1})^{2}$, so we subtract the first term $({ }^{n} C_{0})^{2} = 1$: $({ }^{n} C_{1})^{2} + ({ }^{n} C_{2})^{2} + \ldots + ({ }^{n} C_{n})^{2} = { }^{2n} C_{n} - ({ }^{n} C_{0})^{2} = { }^{2n} C_{n} - 1$.

The value of the sum $\left({ }^{n} C_{1}\right)^{2}+\left({ }^{n} C_{2}\right)^{2}+\left({ }^{n} C_{3}\right)^{2}+\ldots+\left({ }^{n} C_{n}\right)^{2}$ is

Similar Questions

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .......... + C_nx^n$,then $\frac{C_1}{C_0} + \frac{2C_2}{C_1} + \frac{3C_3}{C_2} + .... + \frac{nC_n}{C_{n - 1}} = $

If $C_0, C_1, C_2, \ldots, C_n$ are the binomial coefficients in the expansion of $(1+x)^n$,then the value of $\sum_{r=0}^{n} r^3 \cdot C_r$ when $n=5$ is

$\sum\limits_{r = 0}^m {^{n + r}{C_n} = } $

In the expansion of $(1 + x)^n$,the sum of the coefficients of odd powers of $x$ is:

$\sum_{k=0}^{20} \left({}^{20}C_{k}\right)^{2}$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module