Question

Prove that $[1+(n+1)x]{{(1+x)}^{n-1}}={{C}_{0}}+2.{{C}_{1}}x+3.{{C}_{2}}{{x}^{2}}+\,...\,+(n+1).{{C}_{n}}{{x}^{n}}$ and
${{C}_{0}}^{2}+2.{{C}_{1}}^{2}+3.{{C}_{2}}^{2}+\,......\,+(n+1).{{C}_{n}}^{2}=\dfrac{(n+2)(2n-1)!}{n!(n-1)!}$

Answer 1

Hint: Use the formula${{(1+x)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{0}}x+{}^{n}{{C}_{0}}{{x}^{2}}+\,......\,+{}^{n}{{C}_{n}}{{x}^{n}}$ . Multiply both sides by x , then differentiate to form a new equation. Also, obtain another equation by replacing x in the formula with $\dfrac{1}{x}$. Multiply both the equations and then compare the constant terms of the resultant equation.

Complete step by step answer:
Let us first consider part one of the question.
To prove that $[1+(n+1)x]{{(1+x)}^{n-1}}={{C}_{0}}+2.{{C}_{1}}x+3.{{C}_{2}}{{x}^{2}}+\,...\,+(n+1).{{C}_{n}}{{x}^{n}}$
We know that,
${{(1+x)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+\,...\,+{}^{n}{{C}_{n}}{{x}^{n}}$
Multiplying throughout by $x$, we obtain the equation
$x{{(1+x)}^{n}}={}^{n}{{C}_{0}}x+{}^{n}{{C}_{1}}{{x}^{2}}+{}^{n}{{C}_{2}}{{x}^{3}}+\,...\,+{}^{n}{{C}_{n}}{{x}^{n+1}}...................(1)$
Further we need to differentiate the obtained equation.
The LHS of the equation (1) is in the product form and hence we differentiate LHS by the product rule.
Product rule of Differentiation:
According to product rule the derivative of the product of two terms is equal to the derivative of the first term multiplied by the second term added to the derivative of the second term multiplied by the first term. In our case we take the first term as $x$ and the second term as ${{(1+x)}^{n}}$
Hence, on differentiating both sides of equation (1), we get,
$1{{(1+x)}^{n}}+n{{(1+x)}^{n-1}}x={}^{n}{{C}_{0}}+2{}^{n}{{C}_{1}}x+3{}^{n}{{C}_{2}}{{x}^{2}}+\,...\,+(n+1){}^{n}{{C}_{n}}{{x}^{n}}$
We can see that ${{(1+x)}^{n-1}}$ is common for both the terms of LHS
Hence the equation can be written as,
${{(1+x)}^{n-1}}((1+x)+xn)={}^{n}{{C}_{0}}+2{}^{n}{{C}_{1}}x+3{}^{n}{{C}_{2}}{{x}^{2}}+\,...\,+(n+1){}^{n}{{C}_{n}}{{x}^{n}}$
$[1+(n+1)x]{{(1+x)}^{n-1}}={}^{n}{{C}_{0}}+2{}^{n}{{C}_{1}}x+3{}^{n}{{C}_{2}}{{x}^{2}}+\,......\,+(n+1){}^{n}{{C}_{n}}{{x}^{n}}$
Hence we have proved that $[1+(n+1)x]{{(1+x)}^{n-1}}={}^{n}{{C}_{0}}+2{}^{n}{{C}_{1}}x+3{}^{n}{{C}_{2}}{{x}^{2}}+\,...\,+(n+1){}^{n}{{C}_{n}}{{x}^{n}}$.

Let us now consider part two of the question.
To prove that ${{C}_{0}}^{2}+2.{{C}_{1}}^{2}+3.{{C}_{2}}^{2}+\,......\,+(n+1).{{C}_{n}}^{2}=\dfrac{(n+2)(2n-1)!}{n!(n-1)!}$
We know that,
${{(1+x)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+\,........\,+{}^{n}{{C}_{n}}{{x}^{n}}..............(2)$
In the above equation replacing x with $\dfrac{1}{x}$, we obtain,
${{\left( 1+\dfrac{1}{x} \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}\left( \dfrac{1}{x} \right)+{}^{n}{{C}_{2}}\left( \dfrac{1}{{{x}^{2}}} \right)+\,...\,+{}^{n}{{C}_{n}}\left( \dfrac{1}{{{x}^{n}}} \right)...............(3)$
Multiplying equation (2) throughout by $x$, we obtain the equation
$x{{(1+x)}^{n}}={}^{n}{{C}_{0}}x+{}^{n}{{C}_{1}}{{x}^{2}}+{}^{n}{{C}_{2}}{{x}^{3}}+\,........\,+{}^{n}{{C}_{n}}{{x}^{n+1}}...................(4)$
Now differentiating both sides of equation (4), we get,
$1{{(1+x)}^{n}}+n{{(1+x)}^{n-1}}x={}^{n}{{C}_{0}}+2.{}^{n}{{C}_{1}}x+3.{}^{n}{{C}_{2}}{{x}^{2}}+\,......\,+(n+1).{}^{n}{{C}_{n}}{{x}^{n}}$
$[1+(n+1)x]{{(1+x)}^{n-1}}={}^{n}{{C}_{0}}+2.{}^{n}{{C}_{1}}x+3.{}^{n}{{C}_{2}}{{x}^{2}}+\,.......\,+(n+1).{}^{n}{{C}_{n}}{{x}^{n}}............(5)$
Multiplying equations (3) and (5), we get,
\[\begin{align}
  & {{\left( 1+\dfrac{1}{x} \right)}^{n}}[1+(n+1)x]{{(1+x)}^{n-1}}= \\
 & \left( {}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}\left( \dfrac{1}{x} \right)+{}^{n}{{C}_{2}}\left( \dfrac{1}{{{x}^{2}}} \right)+\,.......\,+{}^{n}{{C}_{n}}\left( \dfrac{1}{{{x}^{n}}} \right) \right)\left( {}^{n}{{C}_{0}}+2.{}^{n}{{C}_{1}}x+3.{}^{n}{{C}_{2}}{{x}^{2}}+\,......\,+(n+1).{}^{n}{{C}_{n}}{{x}^{n}} \right) \\
\end{align}\]
On rearranging,
\[\begin{align}
  & \left( {}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}\left( \dfrac{1}{x} \right)+{}^{n}{{C}_{2}}\left( \dfrac{1}{{{x}^{2}}} \right)+\,...\,+{}^{n}{{C}_{n}}\left( \dfrac{1}{{{x}^{n}}} \right) \right)\left( {}^{n}{{C}_{0}}+2{}^{n}{{C}_{1}}x+3{}^{n}{{C}_{2}}{{x}^{2}}+\,.......\,+(n+1){}^{n}{{C}_{n}}{{x}^{n}} \right) \\
 & ={{\left( 1+\dfrac{1}{x} \right)}^{n}}\left[ 1+(n+1)x \right]{{(1+x)}^{n-1}}...................(6) \\
\end{align}\]

Now collecting constant terms in LHS of the above equation,
Constants (Coefficient of ${{x}^{0}}$) in LHS = ${}^{n}{{C}_{0}}^{2}+2.{}^{n}{{C}_{1}}^{2}+3.{}^{n}{{C}_{2}}^{2}+\,.......\,+(n+1).{}^{n}{{C}_{n}}^{2}$\[RHS={{\left( 1+\dfrac{1}{x} \right)}^{n}}\left[ 1+(n+1)x \right]{{(1+x)}^{n-1}}\]
$=\dfrac{{{\left( 1+x \right)}^{2n-1}}}{{{x}^{n}}}\left[ 1+(n+1)x \right]$
Splitting the terms,
$RHS=\dfrac{{{\left( 1+x \right)}^{2n-1}}}{{{x}^{n}}}+\dfrac{{{\left( 1+x \right)}^{2n-1}}(n+1)}{{{x}^{n-1}}}$
Now considering general terms of the expansions of the above equation,
$RHS=\dfrac{{}^{2n-1}{{C}_{r}}{{x}^{r}}}{{{x}^{n}}}+\dfrac{{}^{2n-1}{{C}_{r}}{{x}^{r}}(n+1)}{{{x}^{n-1}}}$
Collecting only constant terms, we take n = r,
Constants in RHS = ${}^{2n-1}{{C}_{n}}+{}^{2n-1}{{C}_{n}}.(n+1)$
 $=\dfrac{(2n-1)!}{(n-1)!n!}+\dfrac{(2n-1)!(n+1)}{(n-1)!n!}$
$=\dfrac{(2n-1)!(n+2)}{(n-1)!n!}$
Now comparing sum of constant terms in LHS and RHS of equation (6),
${{C}_{0}}^{2}+2.{{C}_{1}}^{2}+3.{{C}_{2}}^{2}+\,......\,+(n+1).{{C}_{n}}^{2}=\dfrac{(n+2)(2n-1)!}{n!(n-1)!}$
Hence proved.

Note:
We consider equating the coefficient of ${{x}^{0}}$ in equation (6) as it will be the only equivalent to the LHS mentioned in the question. None of the other terms’ coefficients will be equal to the required LHS. The formula ${{(1+x)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{0}}x+{}^{n}{{C}_{0}}{{x}^{2}}+\,......\,+{}^{n}{{C}_{n}}{{x}^{n}}$ is a powerful tool, which is very flexible and with simple alteration, is used to obtain several other equation.