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S = \(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{1}{100}.50=\frac{1}{2}\)
Kết luận vậy S > 1/2
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\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+\frac{1}{98\cdot97}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)
\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right)\)
\(\Rightarrow C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(\Rightarrow C=\frac{1}{100}-1+\frac{1}{100}\)
\(\Rightarrow C=\left(\frac{1}{100}+\frac{1}{100}\right)-1\)
\(\Rightarrow C=\frac{1}{50}-1\)
\(\Rightarrow C=\frac{-49}{50}\)
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\(\frac{T}{M}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{1}{99}+\frac{2}{98}+...+\frac{98}{2}+\frac{99}{1}}\)
Xét M - 99 + 98 = \(\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}\)
\(\Leftrightarrow M-1=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)\)
\(\Rightarrow M=\frac{100}{100}+100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(\Rightarrow\frac{T}{M}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}=\frac{1}{100}\)
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1/100.99 - 1/99.98 - 1/98.97 -...- 1/3.2 - 1/2.1
=-(1/100.99 + 1/99.98 + 1/98.97 +...+ 1/3.2 + 1/2.1)
=-(1/2.1+1/3.2 +...+1/98.97+ 1/99.98 +1/100.99 )
=-(1/1.2+1/2.3+1/3.4+...+1/97.98+ 1/98.99 +1/99.100)
=-(1/1-1/2+1/2-1/3+1/3......-1/98+1/98-1/99+1/99-1/100)
=-(1/1-1/100)=-99/100
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B=1/50+1/51+1/52+...+1/99
Ta có: 1/50=1/50
1/51<1/50
1/52<1/50
..............
1/99<1/50
1/50+1/51+1/52+...+1/99<1/50+1/50+1/50+...+1/50(50 phân số 1/50)
B<1
(2 x X - 1)mũ 3 = (2 x X - 2)mũ 2