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A=1x2+2x3+3x4+...+49x50
3A= 3(1.2+2.3+3.4+...+49.50)
3A= 1.2.3+2.3.3+3.4.3+...+49.50.3
3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+49.50.(51-48)
3A= 0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51
3A= 49.50.51
A= 49.50.51/3=41650
B=1x3+3x5+5x7+...+99x101
B=1/1.3 +1/3.5 +...+1/99.101
2B=2/1.3 + 2/3.5 +...+2/99.101
2B=1-1/3+1/3-1/5+...+1/99-1/101
2B=1-1/101
2B=100/101
B=100/101:2=100/202
Lời giải:
$A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{99-97}{97.99}$
$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}$
$=1-\frac{1}{99}=\frac{98}{99}$
\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{98}{99}=\dfrac{1}{33}\cdot49=\dfrac{49}{33}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.........+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
\(=\frac{14}{15}\)
B= 2/3.5 + 2/5.7 +.....+ 2/37.39
= 1/3 - 1/5 + 1/5 - 1/7 + ...... + 1/37 - 1/39
= 1/3 - 1/39 =12/39
C = 1/4 - 1/7 + 1/7 - 1/10 + .....+ 1/73 - 1/ 76
= 1/4 - 1/76 = 18/76
B= 2/3.5 + 2/5.7 +.....+ 2/37.39
= 1/3 - 1/5 + 1/5 - 1/7 + ...... + 1/37 - 1/39
= 1/3 - 1/39 =12/39
C = 1/4 - 1/7 + 1/7 - 1/10 + .....+ 1/73 - 1/ 76
= 1/4 - 1/76 = 18/76
\(M=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(M=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(M=2\left(1-\frac{1}{100}\right)\)
\(M=2.\frac{99}{100}\)
\(M=\frac{99}{50}\)
\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{97.99}\)
\(N=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(N=\frac{3}{2}\left(1-\frac{1}{99}\right)\)
\(N=\frac{3}{2}.\frac{98}{99}\)
\(N=\frac{49}{33}\)
