2xC=2+6+12+20+...+2450

2xC=1x2+2x3+3x4+4x5+...+49x50

6xC=1x2x3+2x3x3+3x4x3+...+49x50x3

6xC=1x2x3+2x3x(4-1)+3x4x(5-2)+...+49x50x(51-48)

6xC=1x2x3+2x3x4-1x2x3+....+49x50x51-48x49x50

6xC=49x50x51

6xC=124950

C=20825

cho mình 1 like nhé

2xC=2+6+12+20+...+2450

2xC=1x2+2x3+3x4+4x5+...+49x50

6xC=1x2x3+2x3x3+3x4x3+...+49x50x3

6xC=1x2x3+2x3x(4-1)+3x4x(5-2)+...+49x50x(51-48)

6xC=1x2x3+2x3x4-1x2x3+....+49x50x51-48x49x50

6xC=49x50x51

6xC=124950

C=20825

k nhé đúng 100%

2C = 2+6+12+20+30+...+2450

2C = 1.2+2.3+3.4+4.5+5.6+...+49.50

3C = 1.2.3+2.3.3+3.4.3+4.5.3+5.6.3+...+49.50.3

3C = 1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+5.6.(7-4)+...+49.50.(51-48)

3C = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+5.6.7-4.5.6+5.6.7-4.5.6+....+49.50.51-48.49.50

=>C=49.50.51 : 3 =>C=4650

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

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

\(2C=2+6+12+20+...+2450\)

\(2C=1.2+2.3+3.4+4.5+...+49.50\)

\(6C=1.2.3+2.3.3+3.4.3+...+49.50.3\)\(6C=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(6C=1.2.3+2.3.4-1.2.3+...+49.50.51-49.50.51\)

\(6C=49.50.51\)

\(6C=124950\)

\(C=20825\)

số các số hạng là:(1225-1):+1=613 (số hạng)

tổng C =(1225+1)x 613:2=375769

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

Google

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}

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

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

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

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

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

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

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

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

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

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

=-2*49/50

=-49/25