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A=1/2 *(1/1*2-1/2*3+1/2*3-1/3*4+........+1/98*99-1/99*100)
=1/2*(1/2-1/99*100)
=1/2*(4950-1/9900)
=4950/19800
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(A=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{99\cdot100}\right]=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
=1+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{2}\) -\(\frac{1}{3}\) -\(\frac{1}{4}\)+\(\frac{1}{3}\) - \(\frac{1}{4}\)-\(\frac{1}{5}\)+.....+\(\frac{1}{99}\)-\(\frac{1}{100}\)-\(\frac{1}{101}\)
=1+\(\frac{1}{101}\)
=\(\frac{102}{101}\)
1/1.2.3 = 1/2 .[1/1.2 - 1 / 2.3]
1/2.3.4 = 1/2[ 1/2- 1/3 ]
...................
1/99.100.101 = 1/2[ 1/99. 100 - 1/100.101]
=> A= 1/2 [ 1/1.2- 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/ 4.5 +.........+ 1/99 .100 - 1/100. 101]
A = 1/2 . [1/1.2 -1/100 .101]
A= 1/2 . 5049 /10100 = 5049 / 20200.
Mình nghĩ là vậy đó.
Biến đổi vế trái ta có:
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+..+\frac{1}{98\cdot99\cdot100}\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99\cdot100}\right)=\frac{4949}{19800}=VP\)
=>đpcm
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{19.20.21}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{20.21}\right)\)
\(=\frac{1}{2}.\frac{209}{420}\)
\(=\frac{209}{840}\)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{19\cdot20\cdot21}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{19\cdot20\cdot21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}-\frac{1}{20\cdot21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot21}\right)\)
bn tự lm tp
\(C=\frac{1.2.3+2.3.4+3.4.5+4.5.6+5.6.7}{3.3.2.3.3.1+2.3.3.3.4.3+3.3.4.3.3.5+3.4.5.3.6.3+3.5.3.6.7.3}+\frac{8}{27}\)
\(C=\frac{1.2.3+2.3.4+3.4.5+4.5.6+5.6.7}{3^3.\left(1.2.3+2.3.4+3.4.5+4.5.6+5.6.7\right)}+\frac{8}{27}\)
\(C=\frac{1}{3^3}+\frac{8}{27}=\frac{1}{27}+\frac{8}{27}=\frac{9}{27}=\frac{1}{3}\)
Vậy C = \(\frac{1}{3}\)