Q40 / 90 · JEE Main 2021 (25 Jul Shift 2) · Mathematics

If $[x]$ be the greatest integer less than or equal to $x,$ then $\sum_{n = 8}^{100} [\frac{{(- 1)}^{n} n}{2}]$ is equal to:

A
$0$
Correct answer
B
$4$
C
$- 2$
D
$2$

Solution

Let $S = \sum_{n = 8}^{100} [\frac{{(- 1)}^{n} \cdot n}{2}]$

Therefore, $S = 4 - 5 + 5 - 6 + 6 + \dots - 50 + 50$

$\Rightarrow S = 4 .$

Hence, the correct answer is option: 'B'.

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