Question

$\mathrm{A}\mathrm{sequence}{x}_1,x_2,x_3,...\mathrm{is}\mathrm{defined}\mathrm{by}\mathrm{letting}{x}_1=2\mathrm{and}{x}_k=\frac{x_{k-1}}{k}\mathrm{for}\mathrm{all}\mathrm{natural}\mathrm{numbers}k,k\ge 2.\mathrm{Show}\mathrm{that}{x}_n=\frac{2}{n!}\mathrm{for}\mathrm{all}n\in \mathbf{N}.$
[NCERT EXEMPLAR]

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{A}\mathrm{sequence}{x}_1,x_2,x_3,...\mathrm{is}\mathrm{defined}\mathrm{by}\mathrm{letting}{x}_1=2\mathrm{and}{x}_k=\frac{x_{k-1}}{k}\mathrm{for}\mathrm{all}\mathrm{natural}\mathrm{numbers}k,k\ge 2. \\ \mathrm{Let}\mathrm{P}\left(n\right):x_n=\frac{2}{n!}\mathrm{for}\mathrm{all}n\in \mathbf{N}. \\ \mathrm{Step}\mathrm{I}:\mathrm{For}n=1, \\ \mathrm{P}\left(1\right):x_1=\frac{2}{1!}=2 \\ \mathrm{So},\mathrm{it}\mathrm{is}\mathrm{true}\mathrm{for}n=1. \\ \mathrm{Step}\mathrm{II}:\mathrm{For}n=k, \\ \mathrm{Let}\mathrm{P}\left(k\right):x_k=\frac{2}{k!}\mathrm{be}\mathrm{true}\mathrm{for}\mathrm{some}k\in \mathbf{N}. \\ \mathrm{Step}\mathrm{III}:\mathrm{For}n=k+1, \\ \mathrm{P}\left(k+1\right): \\ {x}_{k+1}=\frac{x_{k+1-1}}{k} \\ =\frac{x_k}{k} \\ =\frac{2}{k\times k!}\left(\mathrm{Using}\mathrm{step}\mathrm{II}\right) \\ =\frac{2}{\left(k+1\right)!} \\ \mathrm{So},\mathrm{it}\mathrm{is}\mathrm{also}\mathrm{true}\mathrm{for}n=k+1. \\ \mathrm{Hence},x_n=\frac{2}{n!}\mathrm{for}\mathrm{all}n\in \mathbf{N}. \end{gathered}$$

Disclaimer: It should be k instead n in the denominator of $x_k=\frac{x_{k-1}}{k}$. The same has been corrected above.