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In the usual notations prove that
${{C}_{1}}+2{{C}_{2}}x+3{{C}_{3}}{{x}^{2}}+.........n{{C}_{n}}{{x}^{n-1}}=n{{\left( 1+x \right)}^{n-1}}$

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Hint: Notice the pattern and use the binomial expansion of the expression ${{\left( 1+x \right)}^{n}}$ to reach the required result. Differentiation will be needed to reach the desired answer.

Complete step-by-step answer:
Before starting the actual solution, let us try to find the general term for the above series.
So, if we observe it carefully, we will see that the series can be written as ${{C}_{1}}+2{{C}_{2}}x+3{{C}_{3}}{{x}^{2}}+.........n{{C}_{n}}{{x}^{n-1}}=\sum\limits_{r=1}^{n}{r{{C}_{r}}{{x}^{r-1}}}$
Therefore, the general term is ${{T}_{r}}=r{{C}_{r}}{{x}^{r-1}}$
Now let us start with the actual solution to the given equation.
We know that the expansion of ${{\left( 1+x \right)}^{n}}$ is:
${{\left( 1+x \right)}^{n}}=1+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+................{{C}_{n}}{{x}^{n}}$
We know that the derivative of ${{x}^{n}}$ with respect to x is $n{{x}^{n-1}}$ and also the derivative of ${{\left( 1+x \right)}^{n}}$ is $n{{\left( 1+x \right)}^{n-1}}$ . Now we will differentiate both sides of the equation. On doing so, we get
$n{{\left( 1+x \right)}^{n-1}}=0+{{C}_{1}}+2{{C}_{2}}{{x}^{1}}+................n{{C}_{n}}{{x}^{n-1}}$
So, this is equivalent to the equation that we were asked to prove in the question. So, we can say that we have proved the required equation.

Note: Be careful about the signs and try to keep the equations as neat as possible by removing the removable terms. Moreover, make sure that we know the formulas related to different binomial expansions as in case of such questions, they are quite useful. It is prescribed that whenever we see a series including terms of the form $^{n}{{C}_{r}}$ , then always give a thought of using the binomial expansion as this would give us the answer in the shortest possible manner.