-13-13-1100-13-2100-13-2100-13-99100- is equal to- where - denotes greatest integer function

The correct option is B 133 [ 13]+[ 13 1100]+[ 13 2100]+.....+[ 13 99100] =( 1 1 1 ......67 times )+( 2 2 2 ....33 times )= 133 . ...

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Byju's Answer

Standard IV

Mathematics

Addition

[-13]-[-13-11...

Question

$[- \frac{1}{3}] - [- \frac{1}{3} - \frac{1}{100}] + [- \frac{1}{3} - \frac{2}{100}] + . . . + [- \frac{1}{3} + \frac{2}{100}] + . . . . + [- \frac{1}{3} - \frac{99}{100}]$ is equal to? (where $[.]$ denotes greatest integer function)

- 132

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- 133

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- 134

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- 131

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Solution

The correct option is B $- 133$
$[- \frac{1}{3}] + [\frac{- 1}{3} - \frac{1}{100}] + [\frac{- 1}{3} - \frac{2}{100}] + . . . . . + [- \frac{1}{3} - \frac{99}{100}]$
$= (- 1 - 1 - 1 - . . . . . .67$ times $) + (- 2 - 2 - 2 - . . . .33$ times $) = - 133$ .

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Similar questions

For $x \in R, let [x]$ denote the greatest integer $\leq x,$ then the sum of the series $[- \frac{1}{3}] + [- \frac{1}{3} - \frac{1}{100}] + [- \frac{1}{3} - \frac{2}{100}] + . . . . + [- \frac{1}{3} - \frac{99}{100}]$ is :

Q. For

x \in R

, let

[x]

denotes the greatest integer

\leq x

, then the value of

[- \frac{1}{3}] + [- \frac{1}{3} - \frac{2}{100}] + . . . + [- \frac{1}{3} - \frac{99}{100}]

Q. If

a_{1} = 1

and

a_{n + 1} - 3 a_{n} + 2 = 4 n

for every positive integer n, then

a_{100}

equals

Q. If

a_{1} = 1

and

a_{n + 1} - 3 a_{n} + 2 = 4 n

for every positive integer n, then

a_{100}

equals (CAT 2005)

Q. Sum of the series

P = \frac{1}{2 \sqrt{1} + \sqrt{2}} + \frac{1}{3 \sqrt{2} + 2 \sqrt{3}} + . . . . + \frac{1}{100 \sqrt{99} + 99 \sqrt{100}}

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[−13]−[−13−1100]+[−13−2100]+...+[−13+2100]+....+[−13−99100] is equal to? (where [.] denotes greatest integer function)

The correct option is B −133[−13]+[−13−1100]+[−13−2100]+.....+[−13−99100]=(−1−1−1−......67 times )+(−2−2−2−....33 times)=−133.

$[- \frac{1}{3}] - [- \frac{1}{3} - \frac{1}{100}] + [- \frac{1}{3} - \frac{2}{100}] + . . . + [- \frac{1}{3} + \frac{2}{100}] + . . . . + [- \frac{1}{3} - \frac{99}{100}]$ is equal to? (where $[.]$ denotes greatest integer function)

The correct option is B $- 133$
$[- \frac{1}{3}] + [\frac{- 1}{3} - \frac{1}{100}] + [\frac{- 1}{3} - \frac{2}{100}] + . . . . . + [- \frac{1}{3} - \frac{99}{100}]$
$= (- 1 - 1 - 1 - . . . . . .67$ times $) + (- 2 - 2 - 2 - . . . .33$ times $) = - 133$ .