(C) We know that $C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$.The general term of the series is $T_r = \frac{r \cdot C_r}{C_{r-1}}$.Using the property $

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(C) We know that $C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$.The general term of the series is $T_r = \frac{r \cdot C_r}{C_{r-1}}$.Using the property $\frac{C_r}{C_{r-1}} = \frac{n-r+1}{r}$,we get:$T_r = r \cdot \frac{n-r+1}{r} = n-r+1$.The sum is $\sum_{r=1}^{n} (n-r+1) = n + (n-1) + (n-2) + .... + 1$.This is the sum of the first $n$ natural numbers,which is $\frac{n(n+1)}{2}$.

Class 11 Mathematics Binomial Theorem Properties of binomial coefficients

Difficult

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}} = $

A
$\frac{n(n - 1)}{2}$
B
$\frac{n(n + 2)}{2}$
C
$\frac{n(n + 1)}{2}$
D
$\frac{(n - 1)(n - 2)}{2}$

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Binomial Theorem

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Properties of binomial coefficients

If $(1+x-2x^2)^6 = 1+a_1x+a_2x^2+\ldots+a_{12}x^{12}$, then the value of $a_2+a_4+a_6+\ldots+a_{12}$ is

MediumWBJEE 2025

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Let $S_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $S_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
Assertion $(A) : S_3 = 55 \times 2^9$
Reason $(R) : S_1 = 90 \times 2^8$ and $S_2 = 10 \times 2^8$

DifficultAP EAMCET 2025

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If $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$,then $C_0 + 2 C_1 + 3 C_2 + \ldots + (n+1) C_n$ is equal to

DifficultAP EAMCET 2001

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If the coefficients of the $5^{th}$,$6^{th}$,and $7^{th}$ terms in the expansion of $(1 + x)^n$ are in $A.P.$,then $n =$

Medium

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$^nC_0 - \frac{1}{2} ^nC_1 + \frac{1}{3} ^nC_2 - \dots + (-1)^n \frac{^nC_n}{n+1} = $

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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}} = $

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If $(1+x-2x^2)^6 = 1+a_1x+a_2x^2+\ldots+a_{12}x^{12}$, then the value of $a_2+a_4+a_6+\ldots+a_{12}$ is

Let $S_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $S_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.Assertion $(A) : S_3 = 55 \times 2^9$Reason $(R) : S_1 = 90 \times 2^8$ and $S_2 = 10 \times 2^8$

If $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$,then $C_0 + 2 C_1 + 3 C_2 + \ldots + (n+1) C_n$ is equal to

If the coefficients of the $5^{th}$,$6^{th}$,and $7^{th}$ terms in the expansion of $(1 + x)^n$ are in $A.P.$,then $n =$

$^nC_0 - \frac{1}{2} ^nC_1 + \frac{1}{3} ^nC_2 - \dots + (-1)^n \frac{^nC_n}{n+1} = $

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Let $S_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $S_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
Assertion $(A) : S_3 = 55 \times 2^9$
Reason $(R) : S_1 = 90 \times 2^8$ and $S_2 = 10 \times 2^8$