Question

Find its value ${{C}_{1}}+2{{C}_{2}}+3{{C}_{3}}+...+n{{C}_{n}}$ is equal to
(a) ${{2}^{n+1}}$
(b) ${{2}^{n-1}}$
(c) $n{{2}^{n-1}}$
(d) $n{{2}^{n+1}}$

Answer 1

Hint: Use binomial expansion of ${{(a+b)}^{n}}$ where $n$ is a positive integer and $a,b$ are real numbers then ${{(a+b)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{}^{n}{{C}_{n}}{{b}^{n}}$ where ${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}},...,{}^{n}{{C}_{n}}$ are known as binomial coefficients. Put $a=1$ and $b=x$ in the binomial expansion of ${{(a+b)}^{n}}$ . 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 ${{C}_{1}}+2{{C}_{2}}+3{{C}_{3}}+...+n{{C}_{n}}$ . For that we will use binomial expansion of ${{(a+b)}^{n}}$ where $n$ is a positive integer and $a,b$ are real numbers then ${{(a+b)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{}^{n}{{C}_{n}}{{b}^{n}}$ where ${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}},...,{}^{n}{{C}_{n}}$ are known as binomial coefficients. Putting $a=1$ and $b=x$ in the binomial expansion of ${{(a+b)}^{n}}$ we get,
${{(1+x)}^{n}}={}^{n}{{C}_{0}}{{(1)}^{n}}+{}^{n}{{C}_{1}}{{(1)}^{n-1}}x+{}^{n}{{C}_{2}}{{(1)}^{n-2}}{{x}^{2}}+...+{}^{n}{{C}_{n-1}}(1){{x}^{n-1}}+{}^{n}{{C}_{n}}{{x}^{n}}$
After simplification we get,
${{(1+x)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+...+{}^{n}{{C}_{n}}{{x}^{n}}$
For simplification we will be writing ${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}},...,{}^{n}{{C}_{n}}$ as ${{C}_{0}},{{C}_{1}},{{C}_{2}},...,{{C}_{n}}$ so we get,
${{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+...+{{C}_{n}}{{x}^{n}}$
Now we will differentiate the above equation with respect to x, after differentiating we get,
$n{{(1+x)}^{n-1}}={{C}_{1}}+2{{C}_{2}}x+...+n{{C}_{n}}{{x}^{n-1}}........(1)$
Here we have used the derivative formula $\dfrac{d({{x}^{n}})}{dx}=n{{x}^{n-1}}$ 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 ${{C}_{1}}+2{{C}_{2}}+3{{C}_{3}}+...+n{{C}_{n}}$ which is what we have to find in the given question. Therefore putting x=1 in equation (1 )we get,
$\begin{align}
  & n{{(1+1)}^{n-1}}={{C}_{1}}+2{{C}_{2}}(1)+...+n{{C}_{n}}{{(1)}^{n-1}} \\
 & n{{(2)}^{n-1}}={{C}_{1}}+2{{C}_{2}}+...+n{{C}_{n}} \\
\end{align}$
 As whatever be the power on 1 it will always be 1. Thus the value of ${{C}_{1}}+2{{C}_{2}}+3{{C}_{3}}+...+n{{C}_{n}}$ is $n{{2}^{n-1}}$ .

So, the correct answer is “Option C”.

Note: Binomial expansion of ${{(a+b)}^{n}}$ where $n$ is a positive integer and $a,b$ are real numbers then ${{(a+b)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{}^{n}{{C}_{n}}{{b}^{n}}$ where ${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}},...,{}^{n}{{C}_{n}}$ are known as binomial coefficients. Derivative of ${{x}^{n}}$ is given by the formula $\dfrac{d({{x}^{n}})}{dx}=n{{x}^{n-1}}$ .