Question

${lim}_{\pi \to \infty }\frac{1^{99}+2^{99}+3^{99}+\cdots \cdots n^{99}}{n^{100}}=$ [EAMCET 1994]

• A.
  
• B.
  
• C.
  
• D.
  

Answer 1

The correct option is B

${lim}_{\pi \to \infty }\frac{1^{99}+2^{99}+3^{99}+\cdots \cdots n^{99}}{n^{100}}={lim}_{\pi \to \infty }{\sum }_{r=1}^n\left(\frac{r^{99}}{n^{100}}\right)$
$={lim}_{\pi \to \infty }\frac{1}{n}{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^{99}={\int }_0^1{x}^{99}dx={\left[\frac{x^{100}}{100}\right]}_0^1=\frac{1}{100}$