Question

$\left[1+\frac{C_1}{C_0}\right]\left[1+\frac{C_2}{C_1}\right]\left[1+\frac{C_3}{C_2}\right]\dots \dots \left[1+\frac{C_n}{C_{n-1}}\right]=$

• A. $\frac{[n+1]}{n!}$
• B. $\frac{{[n+1]}^n}{(n–1)!}$
• C. $\frac{{[n+1]}^n}{n!}$
• D. $\frac{{[n–1]}^n}{n!}$

Answer 1

The correct option is C

$\frac{{[n+1]}^n}{n!}$

Explanation for the correct option:

Step 1. Expand $C_0$and $C_1$:

Given, $\left[1+\frac{C_1}{C_0}\right]\left[1+\frac{C_2}{C_1}\right]\left[1+\frac{C_3}{C_2}\right]\dots \dots \left[1+\frac{C_n}{C_{n-1}}\right]$

As we know,

$\begin{array}{rcl}{}^{n}C_1& =& \frac{n!}{(n-1)!1!}\\ & =& n\end{array}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}\mathbf{r}\mathbf{!}}\right]$

$\begin{array}{rcl}{}^{n}C_0& =& \frac{n!}{(n-0)!0!}\\ & =& 1\end{array}$

Step 2. Divide ${}^{n}C_1$ by ${}^{n}C_0$, we get

$\frac{C_1}{C_0}=n$

Similarly,

$\begin{array}{rcl}\frac{C_2}{C_1}& =& \frac{{\displaystyle \frac{n!}{\left(n-2\right)!2!}}}{{\displaystyle \frac{n!}{\left(n-1\right)!1!}}}\\ & =& \frac{\left(n-1\right)\left(n-2\right)!}{\left(n-2\right)!2}\\ & =& \frac{\left(n-1\right)}{2}\end{array}$

$\begin{array}{rcl}\frac{C_3}{C_2}& =& \frac{{\displaystyle \frac{n!}{\left(n-3\right)!3!}}}{{\displaystyle \frac{n!}{\left(n-2\right)!2!}}}\\ & =& \frac{\left(n-2\right)\left(n-3\right)!2}{\left(n-3\right)!3\times 2}\\ & =& \frac{\left(n-2\right)}{3}\end{array}$

$\begin{array}{rcl}\frac{C_n}{C_{n-1}}& =& \frac{{\displaystyle \frac{n!}{\left(n-n\right)!n!}}}{{\displaystyle \frac{n!}{\left(n-n+1\right)!\left(n-1\right)!}}}\\ & =& \frac{\left(n-n+1\right)\left(n-1\right)!}{\left(n-n\right)!n!}\\ & =& \frac{\left(n-1\right)!}{n\times \left(n-1\right)!}\\ & =& \frac{1}{n}\end{array}$

Step 3. Put all these values in given expression:

$\left[1+\frac{C_1}{C_0}\right]\left[1+\frac{C_2}{C_1}\right]\left[1+\frac{C_3}{C_2}\right]\dots \dots \left[1+\frac{C_n}{C_{n-1}}\right]=\left[1+n\right]\left[1+\frac{\left(n-1\right)}{2}\right]\left[1+\frac{\left(n-2\right)}{3}\right]...\left[1+\frac{1}{n}\right]$

$=\left(\frac{1+n}{1}\right)\left(\frac{1+n}{2}\right)\left(\frac{1+n}{3}\right)...\left(\frac{1+n}{n}\right)$

$=\frac{{\left(1+n\right)}^{\mathbf{n}}}{\mathbf{n}\mathbf{!}}$ $\left[\mathbf{\because }\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathit{n}\mathbf{=}\mathit{n}\mathbf{!}\right]$

Hence, Option ‘C’ is Correct.