Find the value of -1-8-1-8-1-400-1-8-1-200-1-8-3-400-1-8-399-400- if - x - denotes the integral part of x for real x-A- 75B- 50C- 25D- Cannot be determined

The correct option is B 50As we know that [x]+[ x + 1/n ]+[x+2/n ]+[x+3/n ]+ ⋯ + [x+n 1/n ]=[nx] where n is any integer. Here x=1/8, n = 400. So, sum of the se ...

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Find the value of $[\frac{1}{8}] + [\frac{1}{8} + \frac{1}{400}] + [\frac{1}{8} + \frac{1}{200}] + [\frac{1}{8} + \frac{3}{400}] + \dots + [\frac{1}{8} + \frac{399}{400}]$ if [x] denotes the integral part of x for real x.

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Solution

The correct option is C 50
As we know that
$[x] + [x + \frac{1}{n}] + [x + \frac{2}{n}] + [x + \frac{3}{n}] + \dots + [x + \frac{n - 1}{n}] = [n x]$
where n is any integer. Here $x = \frac{1}{8}$ , n = 400. So, sum of the series is [nx]= $[\frac{400}{8}] = 50$

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Find the value of [18]+[18+1400]+[18+1200]+[18+3400]+⋯+[18+399400] if [x] denotes the integral part of x for real x.

The correct option is C 50As we know that [x]+[x+1n]+[x+2n]+[x+3n]+⋯+[x+n−1n]=[nx] where n is any integer. Here x=18, n = 400. So, sum of the series is [nx]= [4008]=50

Find the value of $[\frac{1}{8}] + [\frac{1}{8} + \frac{1}{400}] + [\frac{1}{8} + \frac{1}{200}] + [\frac{1}{8} + \frac{3}{400}] + \dots + [\frac{1}{8} + \frac{399}{400}]$ if [x] denotes the integral part of x for real x.

The correct option is C 50
As we know that
$[x] + [x + \frac{1}{n}] + [x + \frac{2}{n}] + [x + \frac{3}{n}] + \dots + [x + \frac{n - 1}{n}] = [n x]$
where n is any integer. Here $x = \frac{1}{8}$ , n = 400. So, sum of the series is [nx]= $[\frac{400}{8}] = 50$