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Question: Find its value \({{C}_{1}}+2{{C}_{2}}+3{{C}_{3}}+...+n{{C}_{n}}\) is equal to (a) \({{2}^{n+1}}\)...

Find its value C1+2C2+3C3+...+nCn{{C}_{1}}+2{{C}_{2}}+3{{C}_{3}}+...+n{{C}_{n}} is equal to
(a) 2n+1{{2}^{n+1}}
(b) 2n−1{{2}^{n-1}}
(c) n2n−1n{{2}^{n-1}}
(d) n2n+1n{{2}^{n+1}}

Explanation

Solution

Use binomial expansion of (a+b)n{{(a+b)}^{n}} where nn is a positive integer and a,ba,b are real numbers then (a+b)n=nC0an+nC1an−1b+nC2an−2b2+...+nCn−1abn−1+nCnbn{{(a+b)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{}^{n}{{C}_{n}}{{b}^{n}} where nC0,nC1,nC2,...,nCn{}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}},...,{}^{n}{{C}_{n}} are known as binomial coefficients. Put a=1a=1 and b=xb=x in the binomial expansion of (a+b)n{{(a+b)}^{n}} . Then find the derivative with respect to x and then put the value x=1 and solve.

Complete step by step answer:
In the question we have to find out the value of C1+2C2+3C3+...+nCn{{C}_{1}}+2{{C}_{2}}+3{{C}_{3}}+...+n{{C}_{n}} . For that we will use binomial expansion of (a+b)n{{(a+b)}^{n}} where nn is a positive integer and a,ba,b are real numbers then (a+b)n=nC0an+nC1an−1b+nC2an−2b2+...+nCn−1abn−1+nCnbn{{(a+b)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{}^{n}{{C}_{n}}{{b}^{n}} where nC0,nC1,nC2,...,nCn{}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}},...,{}^{n}{{C}_{n}} are known as binomial coefficients. Putting a=1a=1 and b=xb=x in the binomial expansion of (a+b)n{{(a+b)}^{n}} we get,
(1+x)n=nC0(1)n+nC1(1)n−1x+nC2(1)n−2x2+...+nCn−1(1)xn−1+nCnxn{{(1+x)}^{n}}={}^{n}{{C}_{0}}{{(1)}^{n}}+{}^{n}{{C}_{1}}{{(1)}^{n-1}}x+{}^{n}{{C}_{2}}{{(1)}^{n-2}}{{x}^{2}}+...+{}^{n}{{C}_{n-1}}(1){{x}^{n-1}}+{}^{n}{{C}_{n}}{{x}^{n}}
After simplification we get,
(1+x)n=nC0+nC1x+nC2x2+...+nCnxn{{(1+x)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+...+{}^{n}{{C}_{n}}{{x}^{n}}
For simplification we will be writing nC0,nC1,nC2,...,nCn{}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}},...,{}^{n}{{C}_{n}} as C0,C1,C2,...,Cn{{C}_{0}},{{C}_{1}},{{C}_{2}},...,{{C}_{n}} so we get,
(1+x)n=C0+C1x+C2x2+...+Cnxn{{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+...+{{C}_{n}}{{x}^{n}}
Now we will differentiate the above equation with respect to x, after differentiating we get,
n(1+x)n−1=C1+2C2x+...+nCnxn−1........(1)n{{(1+x)}^{n-1}}={{C}_{1}}+2{{C}_{2}}x+...+n{{C}_{n}}{{x}^{n-1}}........(1)
Here we have used the derivative formula d(xn)dx=nxn−1\dfrac{d({{x}^{n}})}{dx}=n{{x}^{n-1}} for finding the derivative of above equation. We have also used the fact that the derivative of a constant term is zero.
We will put x=1 in the above equation because putting x=1 in the right hand side of equation will turn out to be C1+2C2+3C3+...+nCn{{C}_{1}}+2{{C}_{2}}+3{{C}_{3}}+...+n{{C}_{n}} which is what we have to find in the given question. Therefore putting x=1 in equation (1 )we get,
n(1+1)n−1=C1+2C2(1)+...+nCn(1)n−1 n(2)n−1=C1+2C2+...+nCn \begin{aligned} & n{{(1+1)}^{n-1}}={{C}_{1}}+2{{C}_{2}}(1)+...+n{{C}_{n}}{{(1)}^{n-1}} \\\ & n{{(2)}^{n-1}}={{C}_{1}}+2{{C}_{2}}+...+n{{C}_{n}} \\\ \end{aligned}
As whatever be the power on 1 it will always be 1. Thus the value of C1+2C2+3C3+...+nCn{{C}_{1}}+2{{C}_{2}}+3{{C}_{3}}+...+n{{C}_{n}} is n2n−1n{{2}^{n-1}} .

So, the correct answer is “Option C”.

Note: Binomial expansion of (a+b)n{{(a+b)}^{n}} where nn is a positive integer and a,ba,b are real numbers then (a+b)n=nC0an+nC1an−1b+nC2an−2b2+...+nCn−1abn−1+nCnbn{{(a+b)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{}^{n}{{C}_{n}}{{b}^{n}} where nC0,nC1,nC2,...,nCn{}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}},...,{}^{n}{{C}_{n}} are known as binomial coefficients. Derivative of xn{{x}^{n}} is given by the formula d(xn)dx=nxn−1\dfrac{d({{x}^{n}})}{dx}=n{{x}^{n-1}} .