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5 tháng 5 2015
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\RightarrowĐPCM\)
TP
0
28 tháng 2 2020
\(A=3+3^2+3^3+...+3^{100}\)
\(\Leftrightarrow3A=3^2+3^3+3^4+3^5+....+3^{101}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)
\(\Leftrightarrow2A=3^{101}-3\)
\(\Leftrightarrow A=\frac{3^{101}-3}{2}< 3^{100}-1\)
\(\Leftrightarrow A< B\)
28 tháng 2 2020
a. tính A = 3+3^2+3^3+3^4+.....+3^100
3A=3^2+3^3+3^4+3^5+....+3^100
3A-A=(3^2+3^3+3^4+....+3^101)-(3+3^2+3^3+3^4+.....+3^100)=3^101-3=3^100
mà B=3^100-1 => A<B
NT
Chứng minh rằng:
a) 1/2-1/4+1/8-1/16+1/32-1/64<1/3
b) 1/3-2/3^2+3/3^3-3/3^4+...+99/3^99-100/3^100<3/16
0
N
0
PJ
8
PJ
30 tháng 4 2017
Mình ngu lắm dân trần đăng ninh chuyên anh mà làm sao giỏi toán được
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
=>\(\frac{1}{3}A=\frac{1}{3^2}-\frac{2}{3^3}+\frac{3}{3^4}-\frac{4}{3^5}+....+\frac{99}{3^{100}}-\frac{100}{3^{101}}\)
=>\(\frac{1}{3}A+A=\frac{4}{3}A=\frac{1}{3}-\left(\frac{2}{3^2}-\frac{1}{3^2}\right)+\left(\frac{3}{3^3}-\frac{2}{3^3}\right)+\left(\frac{4}{3^4}-\frac{3}{3^4}\right)+....+\left(\frac{99}{3^{99}}-\frac{98}{3^{99}}\right)+\left(\frac{100}{3^{100}}-\frac{99}{3^{100}}\right)-\frac{100}{3^{101}}\)
=>\(\frac{4}{3}A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+.....+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
Đặt \(S=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
=>\(\frac{1}{3}S=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+....+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
=>\(\frac{1}{3}S+S=\frac{4}{3}S=\frac{1}{3}-\frac{1}{3^{101}}\Rightarrow S=\left(\frac{1}{3}-\frac{1}{3^{101}}\right):\frac{4}{3}=\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{4}=\frac{1}{3}.\frac{3}{4}-\frac{1}{3^{101}}.\frac{3}{4}\)=>\(S=\frac{1}{4}-\frac{1}{3^{100}.4}\)
Mà \(\frac{4}{3}A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+....+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
=>\(\frac{4}{3}A=\frac{1}{4}-\frac{1}{3^{100}.4}=\frac{1}{4}-\frac{1}{3^{100}}.\frac{1}{4}=\frac{1}{4}.\left(1-\frac{1}{3^{100}}\right)\)
=>\(A=\frac{1}{4}\left(1-\frac{1}{3^{100}}\right):\frac{4}{3}=\frac{1}{4}\left(1-\frac{1}{3^{100}}\right).\frac{3}{4}=\frac{1}{4}.\frac{3}{4}.\left(1-\frac{1}{3^{100}}\right)=\frac{3}{16}.\left(1-\frac{1}{3^{100}}\right)\)
Vì \(1-\frac{1}{3^{100}}<1\Rightarrow A<\frac{3}{16}\)