If n is a positive integer,the value of (2n+1 | vedclass

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

Question

(A) The given expression is $S = \sum_{r=0}^n (2n+1-2r) {^nC_r}$.
Expanding the summation:
$S = (2n+1) \sum_{r=0}^n {^nC_r} - 2 \sum_{r=0}^n r \cdot

Vedclass · Accepted Answer

$(n+1) 2^n$

Vedclass · Answer

(A) The given expression is $S = \sum_{r=0}^n (2n+1-2r) {^nC_r}$.
Expanding the summation:
$S = (2n+1) \sum_{r=0}^n {^nC_r} - 2 \sum_{r=0}^n r \cdot {^nC_r}$.
We know that $\sum_{r=0}^n {^nC_r} = 2^n$ and $\sum_{r=0}^n r \cdot {^nC_r} = n \cdot 2^{n-1}$.
Substituting these values:
$S = (2n+1) \cdot 2^n - 2 \cdot (n \cdot 2^{n-1}) = (2n+1) \cdot 2^n - n \cdot 2^n = (2n+1-n) \cdot 2^n = (n+1) \cdot 2^n$.
Also, if $f(x) = x^{n+1}$, then $f'(x) = (n+1)x^n$, so $f'(2) = (n+1)2^n$.
Thus, both options $A$ and $C$ are correct.

If $n$ is a positive integer,the value of $(2n+1) ^nC_0 + (2n-1) ^nC_1 + (2n-3) ^nC_2 + \ldots + 1 \cdot ^nC_n$ is

Similar Questions

Let $X = 1({ }^{10} C _1)^2 + 2({ }^{10} C _2)^2 + 3({ }^{10} C _3)^2 + \ldots + 10({ }^{10} C _{10})^2$,where ${ }^{10} C _{ r }$ for $r \in \{1, 2, \ldots, 10\}$ denotes binomial coefficients. Then,the value of $\frac{1}{1430} X$ is:

$\binom{47}{4} + \sum_{r=1}^5 \binom{52-r}{3} = \dots$

The value of $\frac{1}{1! 50!} + \frac{1}{3! 48!} + \frac{1}{5! 46!} + \dots + \frac{1}{49! 2!} + \frac{1}{51! 1!}$ is $.............$.

For $2 \le r \le n$,$\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$ is equal to

The value of $^{15}C_0^2 - ^{15}C_1^2 + ^{15}C_2^2 - ... - ^{15}C_{15}^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module