Toán 6!(số học)

T

thaotran19

$1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{98}=(1+\dfrac{1}{98})+(\dfrac{1}{2}+\dfrac{1}{97})+...+ (\dfrac{1}{49}+\dfrac{1}{50})$
$=\dfrac{99}{1.98}+\dfrac{99}{2.97}+...+\dfrac{99}{49.50}=99(\dfrac{1}{1.98}+\dfrac{1}{2.97}+...+ \dfrac{1}{49.50})$
\Rightarrow $1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{98}$ chia hết cho 99
\Rightarrow $1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{98}.2.3.4...98$ chia hết cho 99
\Rightarrow $1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{98}.2.3.4...98$ chia cho 99 dư 0
 
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