Question:
For \(x\in\mathbb{R}\), let [x] denote the greatest integer \(\le x\), then the sum of the series
\(\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\left[-\frac{1}{3}-\frac{2}{100}\right]+\ldots+\left[-\frac{1}{3}-\frac{99}{100}\right]\)
is :
For \(x\in\mathbb{R}\), let [x] denote the greatest integer \(\le x\), then the sum of the series
\(\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\left[-\frac{1}{3}-\frac{2}{100}\right]+\ldots+\left[-\frac{1}{3}-\frac{99}{100}\right]\)
is :
\(\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\left[-\frac{1}{3}-\frac{2}{100}\right]+\ldots+\left[-\frac{1}{3}-\frac{99}{100}\right]\)
is :
Show Hint
Each term is \(\left[-\frac13-\frac{k}{100}\right]\), \(k=0,\dots,99\); it equals -1 or -2. Find the switch point.
Updated On: Nov 5, 2026
- - 153
- - 135
- - 133
- - 131
Show Solution
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The Correct Option is C
Solution and Explanation
The k-th term (k = 0 to 99) is \(\left[-\left(\frac13+\frac{k}{100}\right)\right]\).
It equals \(-1\) when \(\frac13+\frac{k}{100}\le1\), i.e. \(k\le66.67\), so \(k=0,1,\dots,66\): 67 terms.
For \(k=67,\dots,99\) we have \(1\lt\frac13+\frac{k}{100}\lt2\) (the maximum is \(0.3333+0.99=1.3233\)), so each term is \(-2\): 33 terms.
Sum \(=67(-1)+33(-2)=-67-66=-133\).
Answer: (3) - 133
It equals \(-1\) when \(\frac13+\frac{k}{100}\le1\), i.e. \(k\le66.67\), so \(k=0,1,\dots,66\): 67 terms.
For \(k=67,\dots,99\) we have \(1\lt\frac13+\frac{k}{100}\lt2\) (the maximum is \(0.3333+0.99=1.3233\)), so each term is \(-2\): 33 terms.
Sum \(=67(-1)+33(-2)=-67-66=-133\).
Answer: (3) - 133
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