(D) We know that $\binom{n}{k} = \binom{n}{n-k}$. Therefore,$\binom{n}{n-r} = \binom{n}{r}$.Substituting this into the expression,we get $\binom{n}{r}

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(D) We know that $\binom{n}{k} = \binom{n}{n-k}$. Therefore,$\binom{n}{n-r} = \binom{n}{r}$.Substituting this into the expression,we get $\binom{n}{r} + \binom{n}{r+1}$.Using the Pascal's identity $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$,the expression simplifies to $\binom{n+1}{r+1}$.

Class 11 Mathematics Binomial Theorem Properties of binomial coefficients

Medium

$\binom{n}{n-r} + \binom{n}{r+1}$,whenever $0 \le r \le n-1$,is equal to

A
$\binom{n}{r-1}$
B
$\binom{n}{r}$
C
$\binom{n}{r+1}$
D
$\binom{n+1}{r+1}$

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Binomial Theorem

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Properties of binomial coefficients

In the expansion of $(1+x)^n$,the sum $\frac{C_1}{C_0} + 2 \frac{C_2}{C_1} + 3 \frac{C_3}{C_2} + \ldots + n \frac{C_n}{C_{n-1}}$ is equal to:

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Let $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. Then the sum $\frac{1}{2^{10}} \sum_{k=0}^{10} \binom{10}{k} k^2$ lies in the interval

KVPY 2021

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

DifficultIIT 2000

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

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$\frac{C_1}{C_0} + 2\frac{C_2}{C_1} + 3\frac{C_3}{C_2} + \dots + 15\frac{C_{15}}{C_{14}} = $

MediumIIT 1962

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$\binom{n}{n-r} + \binom{n}{r+1}$,whenever $0 \le r \le n-1$,is equal to

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In the expansion of $(1+x)^n$,the sum $\frac{C_1}{C_0} + 2 \frac{C_2}{C_1} + 3 \frac{C_3}{C_2} + \ldots + n \frac{C_n}{C_{n-1}}$ is equal to:

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