# Solve the following?

## Let ${\left(1 + x\right)}^{n} = {C}_{0} + {C}_{1} x + {C}_{2} {x}^{2} + \ldots \ldots + {C}_{n} {x}^{n}$ Then the value of $\left(1 + {C}_{0} / {C}_{1}\right) \cdot \left(1 + {C}_{1} / {C}_{2}\right) \ldots . . \cdot \left(1 + {C}_{n - 1} / {C}_{n}\right)$ is?

Apr 15, 2018

(n+1)^n/(n!)

#### Explanation:

$\left(1 + {C}_{0} / {C}_{1}\right) \left(1 + {C}_{1} / {C}_{2}\right) \ldots \left(1 + {C}_{n - 1} / {C}_{n}\right)$

The binomial theorem says that the ${k}^{t h}$ coefficient in the ${n}^{t h}$ order expansion of $\left(1 + x\right)$ is

C_k=(n!)/(k!(n-k)!)

This means that

C_k/C_(k+1)=((k+1)!(n-(k+1))!)/(k!(n-k)!)

=((k+1)!(n-k-1)!)/(k!(n-k)!)=(k+1)/(n-k)

So

$1 + {C}_{k} / {C}_{k + 1} = 1 + \frac{k + 1}{n - k} = \frac{n - k + k + 1}{n - k} = \frac{n + 1}{n - k}$

So our product looks like

((n+1)/(n-0))((n+1)/(n-1))((n+1)/(n-2))...((n+1)/(1))=(n+1)^n/(n!)