Đáp án: A = $\frac{97}{300}$ 

             B = $\frac{101}{306}$

             C = $\frac{200}{101}$ 

             D = $\frac{3}{20}$ 

Giải thích các bước giải:

 a ) A = $\frac{1}{3.4}$ + $\frac{1}{4.5}$ + $\frac{1}{5.6}$ + ..... + $\frac{1}{99.100}$ 

A = $\frac{1}{3}$ - $\frac{1}{4}$ + $\frac{1}{4}$ - $\frac{1}{5}$ + $\frac{1}{5}$ - $\frac{1}{6}$ + ..... + $\frac{1}{99}$ - $\frac{1}{100}$ 

A = $\frac{1}{3}$ - $\frac{1}{100}$ 

A = $\frac{97}{300}$ 

b ) B = $\frac{1}{1.4}$ + $\frac{1}{4.7}$ + $\frac{1}{7.10}$ + ..... + $\frac{1}{99.102}$ 

3 . B = $\frac{3}{1.4}$ + $\frac{3}{4.7}$ + $\frac{3}{7.10}$ + ..... + $\frac{3}{99.102}$ 

3 . B = 1 - $\frac{1}{4}$ + $\frac{1}{4}$ - $\frac{1}{7}$ + $\frac{1}{7}$ - $\frac{1}{10}$ + ..... + $\frac{1}{99.}$ - $\frac{1}{102}$ 

3 . B = 1 - $\frac{1}{102}$ 

3 . B = $\frac{101}{102}$ 

B = $\frac{101}{102}$  : 3

B = $\frac{101}{306}$

c ) C = $\frac{2^2}{1.3}$ + $\frac{2^2}{3.5}$ + $\frac{2^2}{5.7}$ + ..... + $\frac{2^2}{99.101}$ 

C = $\frac{4}{1.3}$ + $\frac{4}{3.5}$ + $\frac{4}{5.7}$ + ..... + $\frac{4}{99.101}$ 

C : 2 = $\frac{2}{1.3}$ + $\frac{2}{3.5}$ + $\frac{2}{5.7}$ + ..... + $\frac{2}{99.101}$ 

C : 2 = 1 - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{5}$ + $\frac{1}{5}$ - $\frac{1}{7}$ + ..... + $\frac{1}{99}$ - $\frac{1}{101}$ 

C : 2 = 1 - $\frac{1}{101}$ 

C : 2 = $\frac{100}{101}$ 

C = $\frac{100}{101}$ . 2

C = $\frac{200}{101}$ 

d ) D = $\frac{1}{2.5}$ + $\frac{1}{5.8}$ + $\frac{1}{8.11}$ + $\frac{1}{11.14}$ + $\frac{1}{14.17}$ + $\frac{1}{17.20}$ 

3 . D = $\frac{3}{2.5}$ + $\frac{3}{5.8}$ + $\frac{3}{8.11}$ + $\frac{3}{11.14}$ + $\frac{3}{14.17}$ + $\frac{3}{17.20}$ 

3 . D = $\frac{1}{2}$ - $\frac{1}{5}$ + $\frac{1}{5}$ - $\frac{1}{8}$ + $\frac{1}{8}$ - $\frac{1}{11}$ + $\frac{1}{11}$ - $\frac{1}{14}$ + $\frac{1}{14}$ - $\frac{1}{17}$ + $\frac{1}{17}$ - $\frac{1}{20}$ 

3 . D = $\frac{1}{2}$ - $\frac{1}{20}$ 

3 . D = $\frac{9}{20}$ 

D = $\frac{9}{20}$  : 3

D = $\frac{3}{20}$ 

$A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}$

$A=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}$

$A=\dfrac{1}{3}-\dfrac{1}{100}=\dfrac{97}{300}$

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$B=\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{99.102}$

$B=\dfrac{1}{3}.(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...\dfrac{1}{99}-\dfrac{1}{102})$

$B=\dfrac{1}{3}.(1-\dfrac{1}{102})$

$B=\dfrac{1}{3}.\dfrac{101}{102}=\dfrac{101}{306}$

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$C=\dfrac{2^2}{1.3}+\dfrac{2^2}{3.5}+...+\dfrac{2^2}{99.101}$

$C=\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{99.101}$

$C=\dfrac{1}{2}.(4-\dfrac{4}{3}+\dfrac{4}{3}-\dfrac{4}{5}+...+\dfrac{4}{99}-\dfrac{4}{101})$

$C=\dfrac{1}{2}.(4-\dfrac{4}{101})$

$C=\dfrac{1}{2}.\dfrac{400}{101}=\dfrac{200}{101}$

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$D=\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{17.20}$

$D=\dfrac{1}{3}.(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20})$

$D=\dfrac{1}{3}.(\dfrac{1}{2}-\dfrac{1}{20})$

$D=\dfrac{1}{3}.\dfrac{9}{20}=\dfrac{3}{20}$