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TM
26 tháng 4 2017
\(S=3+\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^9}\)=> \(2S=6+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^8}\)
=>\(2S-S=\left(6+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^9}\right)\)
=>\(S=3+\frac{3}{2}-\frac{3}{2^9}\) tự tính nốt
LT
2
17 tháng 6 2018
\(3S=3+\frac{1}{3}+...+\frac{1}{3^{2004}}\)
\(3S-S=\left(3+\frac{1}{3}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)
\(2S=3-\frac{1}{3^{2005}}\)
\(2S=\frac{3^{2006-1}}{3^{2005}}\)
\(S=\frac{3^{2006}-1}{3^{2005}.2}\)
17 tháng 6 2018
S = 1/3 + 1/32 + 1/33 + ... + 1/32005
=> 3S = 1 + 1/3 + 1/32 + ... + 1/32004
=> 3S - S = 1 + 1/3 + 1/32 + ... + 1/32004 - (1/3 + 1/32 + 1/33 + ... + 1/32005)
=> 2S = 1 + 1/3 + 1/32 + ... + 1/32004 - 1/3 - 1/32 - 1/33 - ... - 1/32005
=> 2S = 1 - 1/32005
=> S = \(\frac{\frac{1}{3^{2005}}}{2}\)
=> S = 1/32005.2
14 tháng 5 2017
S=\(3\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)
\(S=3\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{5050}\right)\)
\(S=3.\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{10100}\right)\)
\(S=\frac{3}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)
\(S=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(S=\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(S=\frac{3}{2}.\frac{100}{101}=\frac{150}{101}\)
LV
11 tháng 11 2017
\(S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+.....+\frac{n}{2^n}+......+\frac{2017}{2^{2017}}\)
Với n > 2 thì \(\frac{n}{2^n}=\frac{n+1}{2^{n-1}}-\frac{n+2}{2^n}\)
\(\frac{n+1}{2^{n-1}}=\frac{n+1}{2^n:2}=\frac{n+1}{\frac{2^n}{2}}=\frac{2^{\left(n+1\right)}}{2^n}\)
\(\frac{n+1}{2^{n-1}}-\frac{n+2}{2^n}=\frac{2^{n+2}}{2^n}-\frac{n+2}{2^n}\)
\(=\frac{2^{n+2}-n-2}{2^n}\)
\(=\frac{n}{2^n}\)
\(\Leftrightarrow S=\frac{1}{2}+\left(\frac{2+1}{2^{2-1}}-\frac{2+2}{2^2}\right)+.....+\frac{2016+1}{2^{2015}}-\frac{2018}{2^{2016}}\)
\(=\frac{2017+1}{2^{2016}}-\frac{2019}{2^{2017}}\)
\(S=\frac{1}{2}+\frac{3}{2}-\frac{2019}{2017}\)
\(S=2-\frac{2019}{2017}\)
\(\Leftrightarrow S=2-\frac{2019}{2017}< 2\)
Hay \(S< 2\)
24 tháng 6 2018
......................?
mik ko biết
mong bn thông cảm
nha ................
ZG
0
\(2S=6+3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^8}\)
\(=9+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^8}\)
\(\Rightarrow2S-S=\left(9+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)
\(\Rightarrow S=6-\frac{3}{2^9}=6-\frac{3}{512}=\frac{3069}{512}\)