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${{1}^{2}}.{{C}_{1}}+{{2}^{2}}.{{C}_{2}}+{{3}^{2}}.{{C}_{3}}+.........{{n}^{2}}.{{C}_{n}}=n{{\left( 2 \right)}^{n-2}}\left( n+1 \right)$.

Answer
Verified

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 ${{C}_{1}}+{{2}^{2}}.{{C}_{2}}+{{3}^{2}}.{{C}_{3}}+.........{{n}^{2}}.{{C}_{n}}=\sum\limits_{r=1}^{n}{{{r}^{2}}{{C}_{r}}}$

Therefore, the general term is ${{T}_{r}}={{r}^{2}}.{{C}_{r}}$

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

$n{{\left( 1+x \right)}^{n-1}}=0+1.{{C}_{1}}+2{{C}_{2}}{{x}^{1}}+................n{{C}_{n}}{{x}^{n-1}}$

Now multiplying the above equation by x and again differentiating both sides of the equation by x. On doing so, we get

$nx{{\left( 1+x \right)}^{n-1}}=0+1.{{C}_{1}}x+2{{C}_{2}}{{x}^{2}}+................n{{C}_{n}}{{x}^{n}}$

$\Rightarrow n{{\left( 1+x \right)}^{n-1}}+n\left( n-1 \right)x{{\left( 1+x \right)}^{n-2}}=0+1.{{C}_{1}}+{{2}^{2}}.{{C}_{2}}x+................{{n}^{2}}.{{C}_{n}}{{x}^{n-1}}$

Now we will put the value of x=1 in the above equation to prove the asked equation.

$n{{\left( 2 \right)}^{n-1}}+n\left( n-1 \right){{\left( 2 \right)}^{n-2}}=0+1.{{C}_{1}}+{{2}^{2}}.{{C}_{2}}+................{{n}^{2}}.{{C}_{n}}$

$\Rightarrow n{{\left( 2 \right)}^{n-3}}\left( 4+2n-2 \right)=0+1.{{C}_{1}}+{{2}^{2}}.{{C}_{2}}+................{{n}^{2}}.{{C}_{n}}$

$\Rightarrow n{{\left( 2 \right)}^{n-2}}\left( n+1 \right)=0+{{1}^{2}}.{{C}_{1}}+{{2}^{2}}.{{C}_{2}}+................{{n}^{2}}.{{C}_{n}}$

Hence, we have proved the equation given in the question.

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 learn 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 you the answer in the shortest possible manner.

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 ${{C}_{1}}+{{2}^{2}}.{{C}_{2}}+{{3}^{2}}.{{C}_{3}}+.........{{n}^{2}}.{{C}_{n}}=\sum\limits_{r=1}^{n}{{{r}^{2}}{{C}_{r}}}$

Therefore, the general term is ${{T}_{r}}={{r}^{2}}.{{C}_{r}}$

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

$n{{\left( 1+x \right)}^{n-1}}=0+1.{{C}_{1}}+2{{C}_{2}}{{x}^{1}}+................n{{C}_{n}}{{x}^{n-1}}$

Now multiplying the above equation by x and again differentiating both sides of the equation by x. On doing so, we get

$nx{{\left( 1+x \right)}^{n-1}}=0+1.{{C}_{1}}x+2{{C}_{2}}{{x}^{2}}+................n{{C}_{n}}{{x}^{n}}$

$\Rightarrow n{{\left( 1+x \right)}^{n-1}}+n\left( n-1 \right)x{{\left( 1+x \right)}^{n-2}}=0+1.{{C}_{1}}+{{2}^{2}}.{{C}_{2}}x+................{{n}^{2}}.{{C}_{n}}{{x}^{n-1}}$

Now we will put the value of x=1 in the above equation to prove the asked equation.

$n{{\left( 2 \right)}^{n-1}}+n\left( n-1 \right){{\left( 2 \right)}^{n-2}}=0+1.{{C}_{1}}+{{2}^{2}}.{{C}_{2}}+................{{n}^{2}}.{{C}_{n}}$

$\Rightarrow n{{\left( 2 \right)}^{n-3}}\left( 4+2n-2 \right)=0+1.{{C}_{1}}+{{2}^{2}}.{{C}_{2}}+................{{n}^{2}}.{{C}_{n}}$

$\Rightarrow n{{\left( 2 \right)}^{n-2}}\left( n+1 \right)=0+{{1}^{2}}.{{C}_{1}}+{{2}^{2}}.{{C}_{2}}+................{{n}^{2}}.{{C}_{n}}$

Hence, we have proved the equation given in the question.

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 learn 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 you the answer in the shortest possible manner.

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