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Question: Find the value of :- \({C_1} + 2{C_2} + 3{C_3} + ... + n{C_n}\) A.\({2^{n - 1}}\) B.\({2^{n + 1...

Find the value of :- C1+2C2+3C3+...+nCn{C_1} + 2{C_2} + 3{C_3} + ... + n{C_n}
A.2n1{2^{n - 1}}
B.2n+1{2^{n + 1}}
C.n2n1n{2^{n - 1}}
D.n2n+1n{2^{n + 1}}

Explanation

Solution

We will use the formula for the expansion of a binomial term and find the binomial expansion of (1+x)n{\left( {1 + x} \right)^n}. After simplifying it, we will differentiate both sides of the equation with respect to xx. We will use the formula for the differentiation of xn{x^n} with respect to xx to do so. After differentiating the equation, we will substitute 1 for xx in the new equation that we have obtained. After substituting 1 in the new equation, we will obtain the given expression on the right-hand side of the equation and the required answer on the left-hand side of the equation. We will choose the correct option accordingly.
Formulas used:
We will use the following formulas:
1.The binomial expansion of (a+b)n{\left( {a + b} \right)^n} is given by (a+b)n=nC0anb0+nC1an1b1+...+nCna0bn{\left( {a + b} \right)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + ... + {}^n{C_n}{a^0}{b^n}.
2.The differentiation of xn{x^n} with respect to xx is given by ddx(xn)=nxn1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}.

Complete step-by-step answer:
We will calculate the binomial expansion of (1+x)n{\left( {1 + x} \right)^n} by substituting 1 for aa and xx for bb in the formula (a+b)n=nC0anb0+nC1an1b1+...+nCna0bn{\left( {a + b} \right)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + ... + {}^n{C_n}{a^0}{b^n}. Therefore, we get
(1+x)n=nC01nx0+nC11n1x1+nC21n1x2...+nCn10xn{\left( {1 + x} \right)^n} = {}^n{C_0}{1^n}{x^0} + {}^n{C_1}{1^{n - 1}}{x^1} + {}^n{C_2}{1^{n - 1}}{x^2}... + {}^n{C_n}{1^0}{x^n}
Simplifying the above equation, we get
(1+x)n=nC0+nC1x1+nC2x2...+nCnxn\Rightarrow {\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}{x^1} + {}^n{C_2}{x^2}... + {}^n{C_n}{x^n}…………………………………(1)\left( 1 \right)
We will differentiate both sides of equation (1)\left( 1 \right) with respect to xx.
ddx(1+x)n=ddx(nC0+nC1x1+nC2x2...+nCnxn) \Rightarrow \dfrac{d}{{dx}}{\left( {1 + x} \right)^n} = \dfrac{d}{{dx}}\left( {{}^n{C_0} + {}^n{C_1}{x^1} + {}^n{C_2}{x^2}... + {}^n{C_n}{x^n}} \right){\text{ }}
Using the formula ddx(xn)=nxn1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} xx in the above equation, we get
n(1+x)n1=0+nC1+2nC2x+...+nxn1\Rightarrow n{\left( {1 + x} \right)^{n - 1}} = 0 + {}^n{C_1} + 2{}^n{C_2}x + ... + n{x^{n - 1}}
Substituting 1 for xx in the above equation, we get

\Rightarrow n{2^{n - 1}} = {}^n{C_1} + 2{}^n{C_2} + 3{}^n{C_3} + ... + n \\\ $$ The expression on the right-hand side of the equation is the required expression. So, the value of ${C_1} + 2{C_2} + 3{C_3} + ... + n{C_n}$ is $n{2^{n - 1}}$. **$\therefore $ Option C is the correct option.** **Note:** The binomial theorem is not only used to expand expressions of higher powers, but also used to find probability and to solve complex problems of economics, algebra, calculus etc. It is also used in designing infrastructure. We have used the concept of differentiation to simplify the equation further. Differentiate helps to find the derivative of a function with respect to any variable. It measures the rate of change in an independent variable.