If the value of C_{0}+2 cdot C_{1}+3 cdot C_{ | vedclass

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

Question

(A) We are given the sum $S = \sum_{k=0}^{n} (k+1) C_{k} = 576$.We know that $\sum_{k=0}^{n} C_{k} x^{k} = (1+x)^{n}$.Multiplying by $x$,we get $\sum_

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) We are given the sum $S = \sum_{k=0}^{n} (k+1) C_{k} = 576$.We know that $\sum_{k=0}^{n} C_{k} x^{k} = (1+x)^{n}$.Multiplying by $x$,we get $\sum_{k=0}^{n} C_{k} x^{k+1} = x(1+x)^{n}$.Differentiating with respect to $x$: $\sum_{k=0}^{n} (k+1) C_{k} x^{k} = (1+x)^{n} + nx(1+x)^{n-1}$.Setting $x=1$,we get $\sum_{k=0}^{n} (k+1) C_{k} = (1+1)^{n} + n(1)(1+1)^{n-1} = 2^{n} + n \cdot 2^{n-1}$.This simplifies to $2^{n-1}(2+n) = 576$.We can write $576 = 64 	imes 9 = 2^{6} 	imes 9 = 2^{7-1}(7+2)$.Comparing $2^{n-1}(n+2) = 2^{7-1}(7+2)$,we get $n=7$.

If the value of $C_{0}+2 \cdot C_{1}+3 \cdot C_{2}+\ldots+(n+1) \cdot C_{n}=576$,then $n$ is equal to

Similar Questions

Number of rational terms in the expansion of $( \sqrt{2} + \sqrt[4]{3} )^{100}$ is:

If $k$ is the coefficient of $x^5$ in the expansion of $(2x^2 - \frac{1}{3x^3})^5$,then $\frac{3k}{2} = $

In the expansion of $(x-1)(x-2) \ldots (x-18)$, the coefficient of $x^{17}$ is:

In the expansion of $(x^2 - 2x)^{10}$,the coefficient of $x^{16}$ is

If the three consecutive coefficients in the expansion of $(1 + x)^n$ are $28, 56$ and $70,$ then the value of $n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module