Question

find the number of zeros at the end of 5 power n -1 factorial​

Answer 1

Given :

The Factorial of given number : $5^{n-1}$

To find:

We have to find the factorial of $\[{5^{n - 1}}!\]$

Solution :

The factorial of a number is defined as a function that multiplies a number by every number below.

We have to first generalize the pattern,

$\[{5^{1}}!\]$ = 1× 2× 3× 4× 5 = 120, it contains one zero

For a larger number, the number of zeroes can find out by the following method.

$\[{5^{2}}!\]$ =$\[{25}!\]$ = $\[\frac{{25}}{5} + \frac{{25}}{{{5^2}}}\]$ = 6 zeroes

$\[{5^3}!\]$ =  $\[{125}!\]$

    $\[\begin{array}{l} = \frac{{125}}{5} + \frac{{125}}{{{5^2}}} + \frac{{125}}{{{5^3}}}\\\\ = \frac{{{5^3}}}{5} + \frac{{{5^3}}}{{{5^2}}} + \frac{{{5^3}}}{{{5^3}}}\\\\ = 31\end{array}\]$     It contains 31 zeros.

Similarly,

$\[{5^{n - 1}}! = \frac{{{5^{n - 1}}}}{5} + \frac{{{5^{n - 1}}}}{{{5^2}}} + \frac{{{5^{n - 1}}}}{{{5^3}}} + \, \cdot \cdot \cdot \cdot \cdot \cdot \cdot + \frac{{{5^{n - 1}}}}{{{5^{n - 1}}}}\]$

∴ The number of zeros at the end of the $\[{5^{n - 1}}!\]$ = $\[\frac{{{5^{n - 1}}}}{5} + \frac{{{5^{n - 1}}}}{{{5^2}}} + \frac{{{5^{n - 1}}}}{{{5^3}}} + \, \cdot \cdot \cdot \cdot \cdot \cdot \cdot + \frac{{{5^{n - 1}}}}{{{5^{n - 1}}}}\]$  

Answer 2

Step-by-step explanation:

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