For \( m \):
\[ \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \cdots + \frac{1}{\sqrt{99} + \sqrt{100}} = m \]
Rationalize each term:
\[ \frac{\sqrt{1} - \sqrt{2}}{-1} + \frac{\sqrt{2} - \sqrt{3}}{-1} + \cdots + \frac{\sqrt{99} - \sqrt{100}}{-1} = m \]
This telescopes to:
\[ \sqrt{100} - 1 = m \implies m = 9 \]
For \( n \):
\[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{99 \cdot 100} = n \]
Rewrite as:
\[ \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{99} - \frac{1}{100}\right) = n \]
This telescopes to:
\[ 1 - \frac{1}{100} = n \implies n = \frac{99}{100} \]
Thus, \((m, n) = (9, \frac{99}{100})\).
Substitute into the equation:
\[ 11(9) - 100\left(\frac{99}{100}\right) = 99 - 99 = 0 \]
Correct Ans. option (4) \( 11x - 100y = 0 \)
If the sum of the first 10 terms of the series \[ \frac{4 \cdot 1}{1 + 4 \cdot 1^4} + \frac{4 \cdot 2}{1 + 4 \cdot 2^4} + \frac{4 \cdot 3}{1 + 4 \cdot 3^4} + \ldots \] is \(\frac{m}{n}\), where \(\gcd(m, n) = 1\), then \(m + n\) is equal to _____.
If \(\sum\)\(_{r=1}^n T_r\) = \(\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}\) , then \( \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} \) is equal to :
Let \( ABCD \) be a tetrahedron such that the edges \( AB \), \( AC \), and \( AD \) are mutually perpendicular. Let the areas of the triangles \( ABC \), \( ACD \), and \( ADB \) be 5, 6, and 7 square units respectively. Then the area (in square units) of the \( \triangle BCD \) is equal to: