Question

Tính tổng:

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

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

Giúp mik nha, mik đang cần gấp

Answer 1

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

Answer 2

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 

Answer 3

nkky6h njki

Answer 4

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

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

\(=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}\)

\(A=\frac{3}{16}:\frac{1}{2}=\frac{3}{16}.\frac{2}{1}=\frac{3}{8}\)