{C₁} + 2{C₂} + 3{C₃} + 4{C₄} + .... + n{Cₙ} =

English

Gujarati

Hindi

7.Binomial Theorem

medium

${C_1} + 2{C_2} + 3{C_3} + 4{C_4} + .... + n{C_n} = $

A

${2^n}$

B

$n.\,\,{2^n}$

C

$n.\,\,{2^{n - 1}}$

D

$n.\,\,{2^{n + 1}}$

Solution

(c) Trick : Put $n = 1,\,\,2,\,\,3,….$

${S_1} = 1,\,\,{S_2} = 2 + 2 = 4$

Now by alternate $(c)$, put $n = 1,\,\,2$

${S_1} = {1.2^0} = 1,{S_2} = {2.2^1} = 4$

Standard 11

Mathematics

Share

Similar Questions

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

easy

If $n$ is an integer greater than $1$, then $a{ – ^n}{C_1}(a – 1){ + ^n}{C_2}(a – 2) + …. + {( – 1)^n}(a – n) = $

medium

(IIT-1972)

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

medium

(IIT-1962)

A possible value of $^{\prime}x^{\prime}$, for which the ninth term in the expansion of $\left\{3^{\log _{3} \sqrt{25^{x-1}+7}}+3^{\left(-\frac{1}{8}\right) \log _{3}\left(5^{x-1}+1\right)}\right\}^{10}$ in the increasing powers of $3^{\left(-\frac{1}{8}\right) \log _{3}\left(5^{x-1}+1\right)}$ is equal to $180$ , is:

hard

(JEE MAIN-2021)

The value of $^{4n}{C_0}{ + ^{4n}}{C_4}{ + ^{4n}}{C_8} + ….{ + ^{4n}}{C_{4n}}$ is

hard