A fair n (n > 1) faces die is rolled repeatedly until a number less...

Given,

A fair $n\left(n > 1\right)$ faces die is rolled repeatedly until a number less than $n$ appears,

${x}_{i}1234 . . . {p}_{i}\dfrac{n - 1}{n}\dfrac{1}{n} \cdot \left(\dfrac{n - 1}{n}\right)\dfrac{1}{{n}^{2}} \cdot \dfrac{n - 1}{n}\dfrac{1}{{n}^{3}} \cdot \left(\dfrac{n - 1}{n}\right) . . .$

Now we know that, mean is given by,

Mean $= \mathop{∑}_{i = 1}^{\infty}{p}_{i}{x}_{i} = 1 \cdot \dfrac{n - 1}{n} + \dfrac{2}{n} \cdot \left(\dfrac{n - 1}{n}\right) + \dfrac{3}{{n}^{2}}\left(\dfrac{n - 1}{n}\right) + . . .$

$⇒ \dfrac{n}{9} = \left(1 - \dfrac{1}{n}\right)S . . . . . . . . \left(1\right)$

Where, $S = 1 + \dfrac{2}{n} + \dfrac{3}{{n}^{2}} + \dfrac{4}{{n}^{3}} + . . . . . . . . .$\left(2\right)$</p><p>$\dfrac{1}{n}S = \dfrac{1}{n} + \dfrac{2}{{n}^{2}} + \dfrac{3}{{n}^{3}} + . . . . . . . . . \left(3\right)$</p><p>Now subtracting equation \left(2\right) - \left(3\right)$ we get,

\left(1 - \dfrac{1}{n}\right)S = 1 + \dfrac{1}{n} + \dfrac{1}{{n}^{2}} + \dfrac{1}{{n}^{3}} + . . .$</p><p>$⇒ \left(1 - \dfrac{1}{n}\right)S = \dfrac{1}{1 - \dfrac{1}{n}}$</p><p>Now putting the value of$S$in equation \left(1\right)$ we get,

$⇒ \dfrac{n}{9} = \left(1 - \dfrac{1}{n}\right) \times \dfrac{1}{{\left(1 - \dfrac{1}{n}\right)}^{2}} = \dfrac{n}{n - 1}$

$⇒ n = 10$