4N = 1.2.3.4+ 2.3.4.4 + .... + 19.20.21.4

4N = 1.2.3.(4-0) + ...+ 19.20.21.(22-18)

4N = 1.2.3.4 - 0.1.2.3 + .... + 19.20.21.22-18.19.20.21

4N = 19.20.21.22

N = 19.5.21.22

=30076

A=1.2.3+2.3.4+4.5.6+___+19.20.21

4A=1.2.3.4+2.3.4.4+3.4.5.4+___+19.20.21.4

     =1.2.3.(4-0)+2.3.4(5-1)+3.4.5(6-2)+___+19.20.21.(22-18)

     =1.2.3.4-0+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+___+19.20.21.22-19.20.21.18

     =(1.2.3.4-1.2.3.4)+(2.3.4.5-2.3.4.5)+___+(19.20.21.18-19.20.21.18)+19.20.21.22

  A=19.20.21.22:4

A    =43 890

😁

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{19.20.21}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{20.21}\right)\)

\(=\frac{1}{2}.\frac{209}{420}\)

\(=\frac{209}{840}\)

\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{19\cdot20\cdot21}\)

\(=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{19\cdot20\cdot21}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}-\frac{1}{20\cdot21}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot21}\right)\)

bn tự lm tp

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{19.20.21}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{19.20.21}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{19.20}-\frac{1}{20.21}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{20.21}\right)=\frac{1}{2}.\frac{209}{420}=\frac{209}{840}\)

A= \(\frac{1}{1.2.3}\)+ \(\frac{1}{2.3.4}\)+ ... + \(\frac{1}{19.20.21}\)< \(\frac{1}{4}\)

  = 1 - \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{2}\)-  \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ... + \(\frac{1}{19}-\frac{1}{20}-\frac{1}{21}\)

  = 1 - ( \(\frac{1}{2}-\frac{1}{3}\)+ \(\frac{1}{2}-\frac{1}{3}\)) + ... + ( \(\frac{1}{19}-\frac{1}{20}+\frac{1}{19}-\frac{1}{20}\))  - \(\frac{1}{21}\)

  = 1 - \(\frac{1}{21}\)

  =  \(\frac{20}{21}\)<  \(\frac{1}{4}\)

=> Đề bài có sai ko bạn?

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)

\(A=1-\frac{1}{2^{20}}\)

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{21}}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{20}}\)

\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{20}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{21}}\right)\)

\(2B=1-\frac{1}{3^{21}}\)

\(B=\frac{1-\frac{1}{3^{21}}}{2}\)

\(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{19\cdot20\cdot21}\)

\(C=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{19\cdot20\cdot21}\right)\)

\(C=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}-\frac{1}{20\cdot21}\right)\)

\(C=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{20\cdot21}\right)\)

\(C=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{420}\right)\)

\(C=\frac{1}{2}\cdot\frac{209}{420}\)

\(C=\frac{209}{480}\)

\(A=\frac{24}{1.2.3}+\frac{24}{2.3.4}+....+\frac{24}{19.20.21}\)

\(A=24.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{19.20.21}\right)\)

\(A=12.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{19.20.21}\right)\)

\(A=12.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-....-\frac{1}{20.21}\right)\)

\(A=12.\left(\frac{1}{2}-\frac{1}{420}\right)=12.\frac{209}{420}=\frac{209}{35}\)