lim π → ∞ 1 99 + 2 99 + 3 99 + ⋯ ⋯ n 99 n 100 = [EAMCET 1994] Options: A . &nbsp9100 B . &nbsp1100 C . &nbsp199 D . &nbsp1101

Answer: Option B : B lim π → ∞ 1 99 + 2 99 + 3 99 + ⋯ ⋯ n 99 n 100 = lim π → ∞ ∑ n r = 1 ( r 99 n 100 ) = lim π → ∞ 1 n ∑ n r = 1 ( r n ) 99 = ∫ 1 0 x 99 d x = [ x 100 100 ] 1 0 = 1 100

limπ→∞199+299+399+⋯⋯n99n100= [EAMCET 1994]

Question

{lim}_{π \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + \dots \dots n^{99}}{n^{100}} =

[EAMCET 1994]

Options:

A . 9100

B . 1100

C . 199

D . 1101

Answer: Option B
:
B

{lim}_{π \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + \dots \dots n^{99}}{n^{100}} = {lim}_{π \to \infty} \sum_{r = 1}^{n} (\frac{r^{99}}{n^{100}})

= {lim}_{π \to \infty} \frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{99} = \int_{0}^{1} x^{99} d x = {[\frac{x^{100}}{100}]}_{0}^{1} = \frac{1}{100}

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