(n!)/(ncdot (n+1)!+n!cdot (n+1))=(1)/(25) olduguna gôre nka A) 4 B) 5 C) 6 . D) 7

Let's solve the given equation step by step:<br /><br />\[<br />\frac{n!}{n \cdot (n+1)! + n! \cdot (n+1)} = \frac{1}{25}<br />\]<br /><br />First, let's simplify the denominator:<br /><br />\[<br />n \cdot (n+1)! + n! \cdot (n+1)<br />\]<br /><br />We can factor out \(n!\) from both terms:<br /><br />\[<br />n \cdot (n+1)! + n! \cdot (n+1) = n! \cdot (n \cdot (n+1) + (n+1)) = n! \cdot ((n+1)(n+1)) = n! \cdot (n+1)^2<br />\]<br /><br />So the equation becomes:<br /><br />\[<br />\frac{n!}{n! \cdot (n+1)^2} = \frac{1}{25}<br />\]<br /><br />We can cancel out \(n!\) from the numerator and the denominator:<br /><br />\[<br />\frac{1}{(n+1)^2} = \frac{1}{25}<br />\]<br /><br />Now, equate the denominators:<br /><br />\[<br />(n+1)^2 = 25<br />\]<br /><br />Taking the square root of both sides:<br /><br />\[<br />n+1 = \sqrt{25} \quad \text{or} \quad n+1 = -\sqrt{25}<br />\]<br /><br />Since \(n\) must be a positive integer, we only consider the positive root:<br /><br />\[<br />n+1 = 5<br />\]<br /><br />Solving for \(n\):<br /><br />\[<br />n = 5 - 1 = 4<br />\]<br /><br />Therefore, the correct answer is:<br /><br />A) 4