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PH
0
11 tháng 5 2019
Đặt \(S=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3S=1-\frac{2}{3}+\frac{3}{3^2}-...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\\ S+3S=\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)+\left(1-\frac{2}{3}+\frac{3}{3^2}-...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)\\ 4S=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{1}{3^{100}}\\ \Rightarrow12S=3-1+\frac{1}{3}-\frac{1}{3^2}+...-\frac{1}{3^{98}}+\frac{1}{3^{99}}\\ 12S+4S=\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...-\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{1}{3^{100}}\right)\\ 16S=3-\frac{1}{3^{99}}-\frac{1}{3^{99}}-\frac{1}{3^{100}}\\ S=\frac{3-\frac{2}{3^{99}}-\frac{1}{3^{100}}}{16}< \frac{3}{16}\left(đpcm\right)\)
OH
0
PH
0
VP
0
LT
3
9 tháng 5 2016
A=1/3 - 2/3^2+3/3^3 - 4/3^4+ ... - 100/3^100
=>3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99
=>4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100
=>12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99
=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1...
<=>16A=3-101/3^99-100/3^100
<=>A=3/16-(101/3^99+100/3^100)/16 < 3/16
Suy ra A<3/16
A
0
M = 1/3^1 + 2/3^2 + .3/3^3 + .. + 100/3^100
1/3*M= 1/3^2 + 2/3^3 + 3/3^4 + .. + 100/3^101
=> M- 1/3*C = 1/3^1 + (2/3^2 - 1/3^2) + (3/3^3 - 2/3^3) + .. + (100/3^100 - 99/3^100) - 100/3^101
=> 2/3*M = 1/3^1 + 1/3^2 + 1/3^3 + .. + 1/3^100 - 100/3^101
+ xét S= 1/3^1 + 1/3^2 + 1/3^3 + .. + 1/3^100 tương tự
1/3*S = 1/3^2 + 1/3^3 + 1/3^4 + .. + 1/3^101
=> S - 1/3*S = 1/3^1 - 1/3^101
<=> 2/3*S = (1/3 - 1/3^101)
<=> S = 3/2*(1/3 - 1/3^101) thay vào C ta có
2/3*M = 3/2*(1/3 - 1/3^101) - 100/3^101
<> M = 9/4*(1/3 - 1/3^101) - 150/3^101
<>M = 3/4 - 9/4*1/3^101 - 150/3^101 < 3/4
Thấy hay thì tíck cho mk 3 cái
\(M=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\)
\(3M=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3M-M=1+\left(\frac{2}{3}-\frac{1}{3}\right)+\left(\frac{3}{3^2}-\frac{2}{3^2}\right)+...+\left(\frac{100}{3^{99}}-\frac{99}{3^{99}}\right)-\frac{100}{3^{100}}\)
\(2M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(\Rightarrow M=1+\frac{1}{2}=\frac{3}{2}\)
\(\Rightarrow\frac{3}{2}< \frac{3}{4}\left(đpcm\right)\)