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DA
1
NG
0
P
1
1 tháng 1
\(S=\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{2022}}\)
=>\(25\cdot S=1+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2020}}\)
=>\(25S-S=1+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2020}}-\dfrac{1}{5^2}-\dfrac{1}{5^4}-...-\dfrac{1}{5^{2022}}\)
=>\(24S=1-\dfrac{1}{5^{2022}}\)
=>\(S=\dfrac{1}{24}-\dfrac{1}{24\cdot5^{2022}}< \dfrac{1}{24}\)
OP
1
8 tháng 11 2017
S = 1 + 3 + 32 +..+399
=> S = (1 + 3) + ... + (3^98 + 3^99)
=> S = (1 + 3) + ... + 3^98.(1 + 3)
=> S = 4 + ... + 3^98.4
=> S = 4.(1 +... + 3^98) chia hết cho 4 (Đpcm)
HV
0
NT
1
S
11 tháng 5 2017
\(C=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2017}{4^{2017}}\)
\(4C=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2017}{4^{2016}}\)
\(4C-C=\left(1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2017}{4^{2016}}\right)-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2017}{4^{2017}}\right)\)
\(3C=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2016}}-\frac{2017}{4^{2017}}\)
\(12C=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2015}}-\frac{2017}{4^{2016}}\)
\(12C-3C=\left(4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2015}}-\frac{2017}{4^{2016}}\right)-\left(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2016}}-\frac{2017}{4^{2017}}\right)\)
\(9C=4-\frac{2017}{4^{2016}}-\frac{1}{4^{2016}}+\frac{2017}{4^{2017}}\)
\(9C=4-\frac{8068}{4^{2017}}-\frac{4}{4^{2017}}+\frac{2017}{4^{2017}}\)
\(9C=4-\frac{10081}{4^{2017}}\)
=> 9C < 4
=> C < \(\frac{4}{9}\)< \(\frac{1}{2}\)(đpcm)
NM
2
7 tháng 2 2017
thôi để mình làm luôn:
\(C=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+...+2017}\right)\)
\(=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)...\left(1-\frac{1}{\left(2017.2018\right):2}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{\left(2017.2018\right):2-1}{\left(2017.2018\right):2}\)
\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{\left(2017.2018:2-1\right).2}{2017.2018}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{2016.2019}{2017.2018}\)
\(=\frac{\left(1.2...2016\right)\left(4.5...2019\right)}{\left(2.3...2017\right)\left(3.4...2018\right)}\)
\(=\frac{2019}{2017.3}=\frac{2019}{6051}\)
\(S=\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2017}}\)
\(4S=1+\frac{1}{4}+...+\frac{1}{4^{2016}}\)
\(4S-S=\left(1+\frac{1}{4^1}+...+\frac{1}{4^{2016}}\right)-\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2017}}\right)\)
\(3S=1-\frac{1}{4^{2017}}< 1\)
\(\Rightarrow S< \frac{1}{3}\left(đpcm\right)\)