Question

$(1+\frac{C_1}{C_{\circ }})(1+\frac{C_2}{C_1})(1+\frac{C_3}{C_2})..........(1+\frac{Cn}{C_{n-1}})$ is equal to

• A. $\frac{n+1}{n!}$
• B. $\frac{{(n+1)}^n}{{(n-1)}^{!}}$
• C. $\frac{{(n+1)}^n}{n^{!}}$
• D. none of these

Answer 1

The correct option is C $\frac{{(n+1)}^n}{n^{!}}$

$(1+\frac{C_1}{C_0})(1+\frac{C_2}{C_1})(1+\frac{C_3}{C_2})...(1+\frac{C_n}{C_{n+1}})$

As we know that ${{}^{n}C}_r=\frac{n!}{(n-r)!r!}$

So, $C^{}={{}^{n}C}_1=\frac{n!}{(n-1)!1!}=n;C_0={{}^{n}C}_0=\frac{n!}{(n-0)!0!}=1$

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

Similarly expand each term, we will get

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

$(1+n)$ is multiplied by itself $n$ times and

$\mathrm{1.2.3.....}n=n!$

So, answer is $\frac{{(1+n)}^n}{n!}$