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

\(=-2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{2450}\right)\)

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

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

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

B

từ 1 đến 2012 có tất cả:

2012-1:1+1 = 2012 (số)

=>có: 2012:2 = 1006 (cặp)

Mà mỗi cặp bằng (-1)nên

tổng dãy số trên là: 1006 . (-1) = -1006

(1-2)+(2-3)+(3-4)+(5-6)+...+(2011-2012)

=-1+(-1)+(-1)+(-1)+...+(-1)

có tất cả các số -1 trên dãy số trên là

(2012-2);2+1=1006

vậy suy ra ; -1x1006=(-1006)

chac chan la dung

a)A=1/10+1/15+...+1/120

=2(1/20+1/30+...+1/240)

=2(1/4*5+1/5*6+...+1/15*16)

=2*(1/4-1/5+1/5-1/6+...+1/15-1/16)

=2*[(1/4-1/16)+(1/5-1/5)+...+(1/15-1/15)]

=2*[(4/16-1/16)+0+...+0]

=2*3/16=3/8

b) B=1+1/3+1/6+...+1/1225

=2(1/2+1/6+1/12+...+1/2450)

=2(1/1*2+1/2*3+...+1/49*50)

=2*[1-1/2+1/2-1/3+...+1/49-1/50]

=2*[(1-1/50)+(1/2-1/2)+...+(1/49-1/49)]

=2*[(50/50-1/50)+0+...+0]

=2*49/50=49/25

a,\(\frac{1}{2}A=\frac{1}{2}\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)\)

\(\frac{1}{2}A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\)

\(\frac{1}{2}A=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\)

\(\frac{1}{2}A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\)

\(\frac{1}{2}A=\frac{1}{4}-\frac{1}{16}\)\(\frac{1}{2}A=\frac{3}{16}\)suy ra \(A=\frac{3}{16}:\frac{1}{2}=\frac{3}{8}\)

B thì cậu có thể làm nhiều cách 

nkky6h njki

\(B=\left(\frac{1}{3}-1\right).\left(\frac{1}{6}-1\right).\left(\frac{1}{10}-1\right).......\left(\frac{1}{1225}-1\right)\left(\frac{1}{1275}-1\right)\)

\(B=\frac{-2}{3}.\frac{-5}{6}.\frac{-9}{10}......\frac{-1224}{1225}.\frac{-1274}{1275}\)

\(B=\frac{-4}{6}.\frac{-10}{12}.\frac{-18}{20}......\frac{-2448}{2450}.\frac{-2548}{2550}\)

\(B=\frac{-4}{2.3}.\frac{-10}{3.4}.\frac{-18}{4.5}.....\frac{-2448}{49.50}.\frac{-2548}{50.51}\)

\(\Rightarrow\)B có : ( 50 - 2 ) : 1 + 1 = 49 ( số hạng )

\(\Rightarrow B=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}........\frac{2448}{49.50}.\frac{2548}{50.51}.\left(-1\right)\)

     \(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.........\frac{48.51}{49.50}.\frac{49.52}{50.51}.\left(-1\right)\)

     \(B=\frac{\left(1.2.3...48.49\right).\left(4.5.6......51.52\right)}{\left(2.3.4......49.50\right).\left(3.4.5.....50.51\right)}.\left(-1\right)\)

     \(B=\frac{52}{50.3}.\left(-1\right)\)

      \(B=\frac{26}{75}.\left(-1\right)\)

Vậy \(B=\frac{-26}{75}\)