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# Prove the following

Question:

For $\mathrm{x} \in \mathrm{R}$, let $[\mathrm{x}]$ denote the greatest integer $\leq \mathrm{x}$, then the sum of the series

$\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\left[-\frac{1}{3}-\frac{2}{100}\right]+\ldots \ldots+\left[-\frac{1}{3}-\frac{99}{100}\right]$

is

1. $-153$

2. $-133$

3. $-131$

4. $-135$

Correct Option: 2,

Solution:

$\underbrace{\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\ldots+\left[-\frac{1}{3}-\frac{66}{100}\right]}_{(-1) 67}$

$+\underbrace{\left[-\frac{1}{3}-\frac{67}{100}\right]+\ldots+\left[-\frac{1}{3}-\frac{99}{100}\right]}_{-2(33)}=-133$