[Toán 6] bài toán HSG

S

soicon_boy_9x

Ta có:

$\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}
{199}-\dfrac{1}{200}$

$=\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+ \dfrac{ 1 }
{ 200 } -2(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{200}$

$=\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+ \dfrac{ 1 }{ 200 } -
\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{100}$

$=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}
{200}(dpcm)$
 
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