Đáp án:

Giải thích các bước giải:

$\frac{1}{2²} < \frac{1}{1.2} = 1 - \frac{1}{2}$

$\frac{1}{3²} < \frac{1}{2.3} = \frac{1}{2} - \frac{1}{3}$ 

$\frac{1}{4²} < \frac{1}{3.4} = \frac{1}{3} - \frac{1}{4}$

......................................

$\frac{1}{99²} < \frac{1}{98.99} = \frac{1}{98} - \frac{1}{99}$

$\frac{1}{100²} < \frac{1}{99.100} = \frac{1}{99} - \frac{1}{100}$

Cộng tất cả lại:

$\frac{1}{2²} + \frac{1}{3²} + \frac{1}{4²} +...+ \frac{1}{100²} < 1 - \frac{1}{100} < 1$

$C = \frac{3}{4} + \frac{8}{9} + \frac{15}{16} +...+ \frac{9999}{10000}$

$ \frac{2² - 1}{2²} + \frac{3² - 1}{3²} + \frac{4² - 1}{4²} +...+ \frac{100² - 1}{100²}$

$= 1 - \frac{1}{2²} + 1 - \frac{1}{3²} + 1 - \frac{1}{4²} +...+ 1 -  \frac{1}{100²}$

$= 99 - (\frac{1}{2²} + \frac{1}{3²} + 1 + \frac{1}{4²} +...+ \frac{1}{100²})$

$ > 99 - 1 = 98$

C=12²<11.2=1−1212²<11.2=1−12

=13²<12.3=12−1313²<12.3=12−13 

=14²<13.4=13−1414²<13.4=13−14

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=199²<198.99=198−199199²<198.99=198−199

=1100²<199.100=199−11001100²<199.100=199−1100

Cộng tất cả lại:

12²+13²+14²+...+1100²<1−1100<112²+13²+14²+...+1100²<1−1100<1

C=34+89+1516+...+999910000C=34+89+1516+...+999910000

=2²−12²+3²−13²+4²−14²+...+100²−1100²2²−12²+3²−13²+4²−14²+...+100²−1100²

=1−12²+1−13²+1−14²+...+1−1100²=1−12²+1−13²+1−14²+...+1−1100²

=99−(12²+13²+1+14²+...+1100²)=99−(12²+13²+1+14²+...+1100²)

=99−1=98