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Giải:
Ta có: \(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}...+\dfrac{1}{99.100}\)
\(\Leftrightarrow S=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow S=\dfrac{1}{1}-\dfrac{1}{100}\)
\(\Leftrightarrow S=1-\dfrac{1}{100}\)
\(\Leftrightarrow S=\dfrac{99}{100}\)
Vậy ...
1/ tap hop A co 99 phan tu {-1;-2;-3;......;-99}
2/ S=333300
ta có
\(C=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{4.3}+..+\frac{100-99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
1/(1.2)+1/(2.3)+1/(3.4)+...+1/(99.100)
=1-1/2+1/2-1-1/3+1/3-1/4+...+1/99-1/100
=1-1/100
=99/100
tôi không chép bài giang ho đai ca đâu nha.
1/1.2 + 1/2.3 + .................+ 1/99.100 =
1/1 - 1/2 + 1/2 - 1/3 +....................+ 1/99 - 1/100 =
1/1 - 1/100 = 99/100
1/1.2+1/2.3+1/3.4+....+1/99.100
=1/1-1/2+1/2-1/3+1/3-1/4+.....+1/99-1/100
=1/1-1/100
=99/100
\(\frac{3}{4}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{99.100}\)
\(=\frac{3}{4}+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{3}{4}+1-\frac{1}{100}=\frac{87}{50}\)
\(P=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(P=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(P=1-\frac{1}{100}\)
\(P=\frac{99}{100}\)
P = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4+ 1/4 - 1/5 + ............. + 1/99 - 1/100
= 1/1 - 1/100
= 99/100