\(B=-\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{1225}\right)\)

\(\dfrac{1}{2}B=-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{2450}\right)\)

\(\dfrac{1}{2}B=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{2.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\right)\)

\(\dfrac{1}{2}B=-\left(1-\dfrac{1}{50}\right)\)

\(\dfrac{1}{2}B=-1+\dfrac{1}{50}\)

\(\dfrac{1}{2}B=\dfrac{-49}{50}\)

\(B=\dfrac{-49}{25}\)

\(B=-\dfrac{2}{2}-\dfrac{2}{6}-\dfrac{2}{12}-...-\dfrac{2}{2450}\)

\(=-2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{2450}\right)\)

\(=-2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

=-2*49/50

=-49/25

\(B=-1-\frac{1}{3}-\frac{1}{6}-...-\frac{1}{1225}\)

\(=-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)

\(=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(=-2\left(1-\frac{1}{50}\right)=-2\cdot\frac{49}{50}=-\frac{49}{25}\)

\(B=-1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{1225}\)

\(B=-2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\right)\)

\(B=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(B=-2\left(1-\frac{1}{50}\right)\)

\(B=-2\cdot\frac{49}{50}\)

\(B=-\frac{49}{25}\)

D = -1-1/3-1/6-1/10-...-1/1225

Suy ra : D/2=-1/2-1/6-1/12-....-1/2450 
Mà 1/2=1/(1.2)=1-1/2; 1/6=1/(2.3)=1/2-1/3;...1/2450=1/(49.50)=... 
D/2= -(1-1/2)-(-1/2-1/3)-...-(1/49-1/50) 
D/2= -1+1/2-1/2+1/3-....-1/49+1/50 
D/2= -1+1/50=-49/50 
D=(-49/50).2=-98/50

k nha

giúp tôi

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A =  -1-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)-\(\dfrac{1}{10}\)-\(\dfrac{1}{15}\)-...-\(\dfrac{1}{1225}\)

    = -1-(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\))

Đặt B = \(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\)

Ta có : B = 2(\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{2450}\))

               = 2(\(\dfrac{1}{2\text{×}3}\)+\(\dfrac{1}{3\text{×}4}\)+\(\dfrac{1}{4\text{×}5}\)+\(\dfrac{1}{5\text{×}6}\)+...+\(\dfrac{1}{49\text{×}50}\))

               = 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{49}\)-\(\dfrac{1}{50}\)

               = 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{50}\))

               = 2×_{\(\dfrac{24}{50}\)}

                   _{=  \(\dfrac{24}{25}\)}

      _{Thay B vào A ta có :}

   _{A = -1-\(\dfrac{24}{25}\)}

 _{=> A = \(\dfrac{-49}{25}\)}

 _{Cho mik một tick nhé thankss}

\(=-\dfrac{2}{2}-\dfrac{2}{6}-\dfrac{2}{12}-...-\dfrac{2}{2450}\)

\(=-2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(=-2\cdot\dfrac{49}{50}=\dfrac{-49}{25}\)

\(=\dfrac{1\cdot2+2\cdot3+3\cdot4+4\cdot5+...+49\cdot50}{2}\)

\(=\dfrac{49\cdot50\cdot51}{3}=49\cdot50\cdot17=41650\)

Hi