Question

Find the value of $\left( {}^{n}{{C}_{0}} \right)\left( {}^{n}{{C}_{1}} \right)+\left( {}^{n}{{C}_{1}} \right)\left( {}^{n}{{C}_{2}} \right)+.....+\left( {}^{n}{{C}_{n-1}} \right)\left( {}^{n}{{C}_{n}} \right)$.

Answer 1

Hint: We first try to find the binomial expansion of ${{\left( 1+x \right)}^{n}}$. We then try to rearrange this into higher power to lower power terms and change the coefficient of the terms using the method of combination where ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$. Then we take multiplication of those two and find the coefficients of ${{x}^{n-1}}$.

Complete step-by-step answer:
We know that the formula of the binomial expansion is
${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+.....+{}^{n}{{C}_{n-1}}{{x}^{n-1}}+{}^{n}{{C}_{n}}{{x}^{n}}......(i)$.
We can rearrange the terms and start with the maximum power of x first and the form will be
${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{n}}{{x}^{n}}+{}^{n}{{C}_{n-1}}{{x}^{n-1}}+.....+{}^{n}{{C}_{2}}{{x}^{2}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{0}}$.
We change coefficient of the terms using the method of combination where ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$.
${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}+.....+{}^{n}{{C}_{n-2}}{{x}^{2}}+{}^{n}{{C}_{n-1}}x+{}^{n}{{C}_{n}}......(ii)$
Both arrangements have $\left( n+1 \right)$ terms in their expansion.
We take multiplication of those two equations.
$\begin{align}
  & {{\left( 1+x \right)}^{n}}{{\left( 1+x \right)}^{n}} \\
 & =\left( {}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+.....+{}^{n}{{C}_{n-1}}{{x}^{n-1}}+{}^{n}{{C}_{n}}{{x}^{n}} \right)\left( {}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}+.....+{}^{n}{{C}_{n-2}}{{x}^{2}}+{}^{n}{{C}_{n-1}}x+{}^{n}{{C}_{n}} \right) \\
\end{align}$
Now from the long multiplication we try to find the term of x with power $\left( n-1 \right)$.
The power $\left( n-1 \right)$ of x is created from the multiplication of ${{x}^{0}}$ and ${{x}^{n-1}}$, ${{x}^{1}}$ and ${{x}^{n-2}}$, ${{x}^{2}}$ and ${{x}^{n-3}}$, ……., ${{x}^{n-1}}$ and ${{x}^{0}}$.
We only deal with the term of x with power $\left( n-1 \right)$ on both sides.
On the left side the term ${{x}^{n-1}}$ is the ${{n}^{th}}$ term of ${{\left( 1+x \right)}^{n}}{{\left( 1+x \right)}^{n}}={{\left( 1+x \right)}^{2n}}$ which is ${}^{2n}{{C}_{n-1}}{{x}^{n-1}}$.
On the right side we have
$\begin{align}
  & \left( {}^{n}{{C}_{0}}{{x}^{0}} \right)\left( {}^{n}{{C}_{1}}{{x}^{n-1}} \right)+\left( {}^{n}{{C}_{1}}{{x}^{1}} \right)\left( {}^{n}{{C}_{2}}{{x}^{n-2}} \right)+.....+\left( {}^{n}{{C}_{n-1}}{{x}^{n-1}} \right)\left( {}^{n}{{C}_{n}}{{x}^{0}} \right) \\
 & \Rightarrow \left( {}^{n}{{C}_{0}} \right)\left( {}^{n}{{C}_{1}} \right){{x}^{n-1}}+\left( {}^{n}{{C}_{1}} \right)\left( {}^{n}{{C}_{2}} \right){{x}^{n-1}}+.....+\left( {}^{n}{{C}_{n-1}} \right)\left( {}^{n}{{C}_{n}} \right){{x}^{n-1}} \\
 & \Rightarrow \left[ \left( {}^{n}{{C}_{0}} \right)\left( {}^{n}{{C}_{1}} \right)+\left( {}^{n}{{C}_{1}} \right)\left( {}^{n}{{C}_{2}} \right)+.....+\left( {}^{n}{{C}_{n-1}} \right)\left( {}^{n}{{C}_{n}} \right) \right]{{x}^{n-1}} \\
\end{align}$
We equate both sides’ term of x with power $\left( n-1 \right)$.
\[\begin{align}
  & \left[ \left( {}^{n}{{C}_{0}} \right)\left( {}^{n}{{C}_{1}} \right)+\left( {}^{n}{{C}_{1}} \right)\left( {}^{n}{{C}_{2}} \right)+.....+\left( {}^{n}{{C}_{n-1}} \right)\left( {}^{n}{{C}_{n}} \right) \right]{{x}^{n-1}}={}^{2n}{{C}_{n-1}}{{x}^{n-1}} \\
 & \Rightarrow \left( {}^{n}{{C}_{0}} \right)\left( {}^{n}{{C}_{1}} \right)+\left( {}^{n}{{C}_{1}} \right)\left( {}^{n}{{C}_{2}} \right)+.....+\left( {}^{n}{{C}_{n-1}} \right)\left( {}^{n}{{C}_{n}} \right)={}^{2n}{{C}_{n-1}} \\
\end{align}\]
So, the coefficients are same and the value of $\left( {}^{n}{{C}_{0}} \right)\left( {}^{n}{{C}_{1}} \right)+\left( {}^{n}{{C}_{1}} \right)\left( {}^{n}{{C}_{2}} \right)+.....+\left( {}^{n}{{C}_{n-1}} \right)\left( {}^{n}{{C}_{n}} \right)$ is ${}^{2n}{{C}_{n-1}}{{x}^{n-1}}$.

Note: We need to remember solving the general term of the series $\left( {}^{n}{{C}_{0}} \right)\left( {}^{n}{{C}_{1}} \right)+\left( {}^{n}{{C}_{1}} \right)\left( {}^{n}{{C}_{2}} \right)+.....+\left( {}^{n}{{C}_{n-1}} \right)\left( {}^{n}{{C}_{n}} \right)$ and then find the summation will never solve it. The coefficients are broken into particular patterns and that can only be solved by multiplication of the expansions. There won't be any other coefficients of ${{x}^{n-1}}$ other than those we have already taken into account.