_ n 1 n _ r = 1 ^ 2n r n^2 + r^2 equals - Vidyadip

MCQ

$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^{2n} {\frac{r}{{\sqrt {{n^2} + {r^2}} }}} $ equals

A
$1 + \sqrt 5 $
✓
$ - 1 + \sqrt 5 $
C
$ - 1 + \sqrt 2 $
D
$1 + \sqrt 2 $

✓

Answer

Correct option: B.

$ - 1 + \sqrt 5 $

b
(b) $L = \mathop {\lim }\limits_{n \to \infty } \,\,\sum\limits_{r = 1}^{2n} {} \frac{1}{n}\,.\,\frac{{r/n}}{{\sqrt {1 + {{(r/n)}^2}} }}$

$= \int_0^2 {} \frac{x}{{\sqrt {1 + {x^2}} }}\,dx = \sqrt 5 - 1$.

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