CMR : 1/2! + 2/3! + 3/4! + ... + 99/100! < 1
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BT
0
KS
0
6 tháng 4 2016
100=10*10
100=1000:10
100 câu nói hay về cuộc sống
LN
0
NV
0
NV
17
LC
6 tháng 1 2016
Đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
=>\(\frac{1}{3}.A=\frac{1}{3^2}-\frac{2}{3^3}+\frac{3}{3^4}-\frac{4}{3^5}+...+\frac{99}{3^{100}}-\frac{100}{3^{101}}\)
=>\(A+\frac{1}{3}.A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}+\frac{1}{3^2}-\frac{2}{3^3}+\frac{3}{3^4}-\frac{4}{3^5}+...+\frac{99}{3^{100}}-\frac{100}{3^{101}}\)
=>\(\frac{4}{3}.A=\frac{1}{3}-\left(\frac{2}{3^2}-\frac{1}{3^2}\right)+\left(\frac{3}{3^3}-\frac{2}{3^3}\right)-\left(\frac{4}{3^4}-\frac{3}{3^4}\right)+...+\left(\frac{99}{3^{99}}-\frac{98}{3^{99}}\right)-\left(\frac{100}{3^{100}}-\frac{99}{3^{100}}\right)-\frac{100}{3^{101}}\)
=>\(\frac{4}{3}.A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}-\frac{100}{3^{101}}\)
Đặt \(B=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
=>\(\frac{1}{3}.B=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
=>\(B+\frac{1}{3}.B=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
=>\(\frac{4}{3}.B=\frac{1}{3}-\frac{1}{3^{101}}\)
=>\(B=\frac{1}{3}:\frac{4}{3}-\frac{1}{3^{101}}:\frac{4}{3}\)
=>\(B=\frac{1}{3}.\frac{3}{4}-\frac{1}{3^{101}}.\frac{3}{4}\)
=>\(B=\frac{1}{4}-\frac{1}{3^{100}.4}\)
Lại có: \(\frac{4}{3}.A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}-\frac{100}{3^{101}}\)
=>\(\frac{4}{3}.A=B-\frac{100}{3^{101}}\)
=>\(\frac{4}{3}.A=\frac{1}{2}-\frac{1}{3^{100}.4}-\frac{100}{3^{101}}\)
=>\(\frac{4}{3}.A=\frac{1}{2}-\left(\frac{1}{3^{100}.4}+\frac{100}{3^{101}}\right)\)
=>\(\frac{4}{3}.A=\frac{1}{2}-\left(\frac{1}{3^{100}}.\frac{1}{4}+\frac{1}{3^{100}}.\frac{100}{3}\right)\)
=>\(\frac{4}{3}.A=\frac{1}{2}-\frac{1}{3^{100}}.\left(\frac{1}{4}+\frac{100}{3^{ }}\right)\)
=>\(\frac{4}{3}.A=\frac{1}{2}-\frac{1}{3^{100}}.\frac{403}{12}\)
Ta thấy: \(\frac{1}{3^{100}}.\frac{403}{12}<\frac{1}{3}.\frac{9}{12}=\frac{1}{3}.\frac{3}{4}=\frac{1}{4}\)
=>\(\frac{1}{3^{100}}.\frac{403}{12}<\frac{1}{4}\)
=>\(\frac{4}{3}.A=\frac{1}{2}-\frac{1}{3^{100}}.\frac{403}{12}<\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)
=>\(\frac{4}{3}.A<\frac{1}{4}=>A<\frac{1}{4}:\frac{4}{3}=>A<\frac{3}{16}\)
=>\(A<\frac{3}{16}\)
Vậy \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)
EB
0
NV
0
NT
0
12 tháng 5 2020
\(=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
Ta có :
1/2! + 2/3! + 3/4! + ... + 99/100!
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)( đpcm )