1+C1C01+C2C11+C3C2……1+CnCn-1 is equal to

1+C1C0 1+C2C1 1+C3C2 ......... 1+CnCn-1 =1+n-1+11 1+n-2+12 1+n-3+13......1+n-n+1n =(n+1)1 (n+1)2 (n+1)3 ..... n+1n =(n+1)nn !

1+C1C01+C2C11+C3C2……1+CnCn-1 is equal to

1+C1C0 1+C2C1 1+C3C2 ......... 1+CnCn-1 =1+n-1+11 1+n-2+12 1+n-3+13......1+n-n+1n =(n+1)1 (n+1)2 (n+1)3 ..... n+1n =(n+1)nn !

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$(1 + \frac{C_{1}}{C_{0}}) (1 + \frac{C_{2}}{C_{1}}) (1 + \frac{C_{3}}{C_{2}}) \dots \dots (1 + \frac{C_{n}}{C_{n - 1}})$ is equal to

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$\frac{{(n - 1)}^{n}}{n!}$

$\frac{{(n + 1)}^{n}}{n!}$

$\frac{{(n + 1)}^{n}}{(n - 1)!}$

$\frac{n + 1}{n!}$

answer is A.

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Detailed Solution

$\begin{array}{rcl} (1 + \frac{C_{1}}{C_{0}}) (1 + \frac{C_{2}}{C_{1}}) (1 + \frac{C_{3}}{C_{2}}) . . . . . . . . . (1 + \frac{C_{n}}{C_{n - 1}}) \\ = & (1 + \frac{n - 1 + 1}{1}) (1 + \frac{n - 2 + 1}{2}) (1 + \frac{n - 3 + 1}{3}) . . . . . . (1 + \frac{n - n + 1}{n}) \\ = & \frac{(n + 1)}{1} \frac{(n + 1)}{2} \frac{(n + 1)}{3} . . . . . (\frac{n + 1}{n}) \\ = & \frac{{(n + 1)}^{n}}{n!} \end{array}$