Câu trắc nghiệm trả lời ngắn

Đáp số: 2500.

$1 + 3 + 5 + \ldots + 99 = \frac{{(1 + 99) \cdot 50}}{2} = 2500$

Đáp số: 496.

$\sin {\rm{x}} = 0 \Leftrightarrow {\rm{x}} = {\rm{k}}\pi ,{\rm{k}} \in \mathbb{Z},0 \le {\rm{k}}\pi \le 100 \Leftrightarrow {\rm{k}} \le \frac{{100}}{\pi },\frac{{100}}{\pi } \approx 31{,}8$

$\Rightarrow {\rm{k}}\pi \in 0;\pi ;2\pi ;3\pi ; \ldots ;31\pi$

$0 + \pi + 2\pi + 3\pi + \ldots + 31\pi = \frac{{(\pi + 31\pi ) \cdot 31}}{2} = 496\pi$

Đáp số: -1.

Ta có $\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \frac{1}{{{2^4}}} + \ldots + \frac{1}{{{2^{30}}}} = \frac{{\frac{1}{2} \cdot \left( {1 - \frac{1}{{{2^{30}}}}} \right)}}{{1 - \frac{1}{2}}} = 1 - \frac{1}{{{2^{30}}}} \Rightarrow a \cdot b = - 1$

Đáp số: 1.

$\frac{{1 + 3 + 5 + 7 + \ldots + (2n - 1)}}{{{{\rm{n}}^2}}} = \frac{{\frac{{[1 + (2n - 1)] \cdot n}}{2}}}{{{{\rm{n}}^2}}} = \frac{{{{\rm{n}}^2}}}{{{{\rm{n}}^2}}} = 1$