Lời giải:
17)
$A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{99.100}$
$A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$A=\dfrac{1}{1}-\dfrac{1}{100}$
`A=99/100`
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18)
$A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.10}+...\dfrac{1}{97.99}$
$2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{99}$
$2A=\dfrac{1}{3}-\dfrac{1}{99}$
$2A=\dfrac{32}{99}$
$A=\dfrac{32}{99}÷2$
$A=\dfrac{16}{99}$
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19)
$S=(-3)^0+(-3)^1+(-3)^2+...+(-3)^{2004}$
$-3S=(-3)^1+(-3)^2+(-3)^3+...+(-3)^{2005}$
$-3S-S=[(-3)^1+(-3)^2+(-3)^3+...+(-3)^{2005}]-[(-3)^0+(-3)^1+(-3)^2+...+(-3)^{2004}]$
$-4S=(-3)^{2005}-1$
$S=\dfrac{(-3)^{2005}-1}{4}$
