Question

Let $ X={{\left( ^{10}{{C}_{1}} \right)}^{2}}+{{\left( ^{10}{{C}_{1}} \right)}^{2}}+3{{\left( ^{10}{{C}_{1}} \right)}^{2}}...+10{{\left( ^{10}{{C}_{1}} \right)}^{2}}$, where $^{10}{{C}_{r}},r=\{1,2,3,...,10\}$ denote binomial coefficients .Then find the value of $\dfrac{1}{1430}X$.

Answer 1

Hint: Convert the given expression to generalized form to apply the standard formula of $^{2n-1}{{C}_{n-1}}$. Then make the given data a specialized case of this to find out the required value.

Complete step-by-step answer:
The given expression can written in generalized form $X={{\sum\limits_{r=0}^{n}{r\left( ^{n}{{C}_{r}} \right)}}^{2}}=\sum\limits_{r=0}^{n}{r\left( ^{n}{{C}_{r}} \right)}\left( ^{n}{{C}_{r}} \right)$.
 We can replace $^{n}{{C}_{r}}=\dfrac{n}{r}\left( ^{n-1}{{C}_{r-1}} \right)$ . Now the expression transforms to
$\begin{align}
  & X={{\sum\limits_{r=0}^{n}{r\left( ^{n}{{C}_{r}} \right)}}^{2}} \\
 & \Rightarrow X=\sum\limits_{r=0}^{n}{r}\left( ^{n}{{C}_{r}} \right)\left( ^{n}{{C}_{r}} \right) \\
 & \Rightarrow X=\sum\limits_{r=0}^{n}{r\left( ^{n}{{C}_{r}} \right)}\left( ^{n-1}{{C}_{r-1}} \right)\left( \dfrac{n}{r} \right) \\
 & \Rightarrow X=n\sum\limits_{r=0}^{n}{\left( ^{n}{{C}_{r}} \right)}\left( ^{n-1}{{C}_{r-1}} \right) \\
\end{align}$\[\]
We use the fact that $^{n}{{C}_{r}}{{=}^{n}}{{C}_{n-r}}$\[\]
We also know from theory of binomial expansion that $\sum\limits_{r=0}^{n}{\left( ^{n}{{C}_{n-r}} \right)}\left( ^{n-1}{{C}_{r-1}} \right){{=}^{2n-1}}{{C}_{n-1}}$. Putting it in above equation \[\]
$\begin{align}
  & X=n\sum\limits_{r=0}^{n}{\left( ^{n}{{C}_{r}} \right)}\left( ^{n-1}{{C}_{r-1}} \right) \\
 & \Rightarrow X=n\sum\limits_{r=0}^{n}{\left( ^{n}{{C}_{n-r}} \right)}\left( ^{n-1}{{C}_{r-1}} \right) \\
 & \Rightarrow X=n\left( ^{2n-1}{{C}_{n-1}} \right) \\
\end{align}$\[\]
Now we apply for the special case as asked in the question . So
\[X=10\cdot \left( ^{20-1}{{C}_{10-1}} \right)=10\left( ^{19}{{C}_{10}} \right)\]

 We have been asked to find out the value of $\dfrac{1}{1430}X$. So we first factorize 1430 as $1430=10.11.13$. Using this obtained value to substitute in the above equation.
\[\dfrac{1}{1430}X=10\dfrac{19\times 18\times ...11}{9\times 8\times ...2\times 10\times 11\times 13}=646\]
The required value is 646.\[\]

Note: We need to be careful of wrong substitution as it may lead to incorrect results. We need to be also careful of the fact that the question is asking the values of $\dfrac{1}{1430}X$ not $X$. So do not end the solution at $X$ .