Question

A sequence $x_1,x_2,x_3,......$ is defined by letting $x_1=2$ and $x_k=\frac{x_{k-1}}{n}$ for all natural numbers k, $k\ge 2$. Show that $x_n=\frac{2}{n!}$ for all $nϵN$.

Answer 1

Let P(n) : $x_n=\frac{2}{n!}$, $\forall \cap ϵN$ to prove P(2) is true.

Step 1: P(2) : $x_2=\frac{2}{2!}=\frac{2}{2\times 1}=1$

As, given $x_1=2$

$x_k=\frac{x_{k-1}}{k}$

$x_2=\frac{x_1}{2}=\frac{2}{2}=1$

Hence, P(2) is true.

Step II : Now, assume that P(k) is true.

$P\left(k\right):x_k=\frac{2}{k!}$

Step III : Now, to prove that P(k + 1) is true, we have to show that

$P(k+1):x_{k+1}=\frac{2}{(k+1)}$

$x_{k+1}=\frac{x_{k+1-1}}{k}=\frac{x_k}{k}$

$=\frac{2}{k!k}=\frac{2}{(k+1)!}$

So, P(k + 1) is true. Hence, P(n) is true.