Question

find the value lt x=x (1/x²+ 2/x²+3/x²....x/x²)​

Answer 1

Answer:

x = 1

⇒ $x(\frac{1}{x^2} + \frac{2}{x^2} + \frac{3}{x^2} +...+\frac{x}{x^2} ) = x = 1$

Step-by-step explanation:

Given :

To find the value of x,

if $x = x(\frac{1}{x^2} + \frac{2}{x^2} + \frac{3}{x^2} +...+\frac{x}{x^2} )$

Solution :

$x = x(\frac{1}{x^2} + \frac{2}{x^2} + \frac{3}{x^2} +...+\frac{x}{x^2} )$

Dividing both the sides by x ,

we get,

⇒ $x(\frac{1}{x}) = (\frac{1}{x})x(\frac{1}{x^2} + \frac{2}{x^2} + \frac{3}{x^2} +...+\frac{x}{x^2} )$

⇒ $1 = \frac{1}{x^2} + \frac{2}{x^2} + \frac{3}{x^2} +...+\frac{x}{x^2}$

As, all the terms have same denominator (no need to coss-multiply/find LCM)

⇒ $1 = \frac{1+2+3+...+x}{x^2}$

We know that,

⇒ $1 + 2 + 3 + 4+ ... + n = \frac{n(n+1)}{2}$

Hence,

⇒ $1 + 2 + 3 + 4+ ... + x = \frac{x(x+1)}{2}$

So,

⇒ $1 = \frac{\frac{x(x+1)}{2}}{x^2} = \frac{x(x+1)}{2x^2}=\frac{x+1}{2x}$

⇒ $1 = \frac{x+1}{2x}$

⇒ 2x = x + 1

⇒ 2x - x = 1

⇒ x = 1

So,

⇒ $x(\frac{1}{x^2} + \frac{2}{x^2} + \frac{3}{x^2} +...+\frac{x}{x^2} ) = x = 1$