S= 1/2 + 1/2^2 + 1/2^3 + .......1/2^2014
a/ tính tổng S
b/ CMR: S <1
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S
1 tháng 5 2017
\(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+....+\frac{2014}{4^{2014}}\)
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2014}{4^{2013}}\)
\(4S-S=\left(1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2014}{4^{2013}}\right)-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2014}{4^{2014}}\right)\)
\(3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
\(12S=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}-\frac{2014}{4^{2013}}\)
\(12S-3S=\left(4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}-\frac{2014}{4^{2013}}\right)-\left(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\right)\)
\(9S=4-\frac{2014}{4^{2013}}-\frac{1}{4^{2013}}+\frac{2014}{4^{2014}}\)
\(9S=4-\frac{4028}{4^{2014}}-\frac{4}{4^{2014}}+\frac{2014}{4^{2014}}\)
\(9S=4-\frac{2010}{4^{2014}}< 4\)
\(\Rightarrow9S< 4\)
\(\Rightarrow S< \frac{4}{9}< 1\)(đpcm)
1 tháng 5 2017
Ta có :
\(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\)( 1 )
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2014}{4^{2013}}\)( 2 )
Lấy ( 2 ) - ( 1 ) ta được :
\(3S=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
gọi \(B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}\)( 3 )
\(4B=4+1+\frac{1}{4}+...+\frac{1}{4^{2012}}\) ( 4 )
Lấy ( 4 ) - ( 3 ) ta được :
\(3B=4-\frac{1}{4^{2013}}\)
\(\Rightarrow B=\frac{4-\frac{1}{4^{2013}}}{3}=\frac{4}{3}-\frac{1}{4^{2013}.3}\)
\(\Rightarrow3S=\frac{4}{3}-\frac{1}{4^{2013}.3}-\frac{2014}{4^{2014}}\)
\(\Rightarrow S=\frac{\frac{4}{3}-\frac{1}{4^{2013}.3}-\frac{2014}{4^{2014}}}{3}=\frac{4}{9}-\frac{1}{4^{2013}.9}-\frac{2014}{4^{2014}.3}< \frac{4}{9}< 1\)
vậy \(S< 1\)
HH
1
15 tháng 4 2017
R.2=2+2^2+2^3+...+2^2015
R=(2+2^2+2^3+...+2^2015-1)-(1+2^2+2^3+...+2^2014)
R=(2^2015)-1
S=(2^2015)-1 / 1-(2^2015)
S=-1
bieu thuc do goi la R nhe
DD
Đoàn Đức Hà
Giáo viên
27 tháng 5 2021
\(S=2014+\frac{2014}{1+2}+\frac{2014}{1+2+3}+...+\frac{2014}{1+2+3+...+10000}\)
\(S=\frac{2014}{\frac{1.2}{2}}+\frac{2014}{\frac{2.3}{2}}+\frac{2014}{\frac{3.4}{2}}+...+\frac{2014}{\frac{10000.10001}{2}}\)
\(S=\frac{4028}{1.2}+\frac{4028}{2.3}+\frac{4028}{3.4}+...+\frac{4028}{10000.10001}\)
\(S=4028\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10000.10001}\right)\)
\(S=4028\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10001-10000}{10000.10001}\right)\)
\(S=4028\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10000}-\frac{1}{10001}\right)\)
\(S=4028\left(1-\frac{1}{10001}\right)=\frac{40280000}{10001}\)
NA
2
29 tháng 1 2023
\(\cdot DuyNam\)
\(S=\left(-1\right)+\left(-1\right)^2+\left(-1\right)^3+...+\left(-1\right)^{2014}+\left(-1\right)^{2015}\)
\(S=\left(-1\right)+1+\left(-1\right)+...+1+\left(-1\right)\) (2015 thừa số)
`-> S= (-1)`
P
6
14 tháng 7 2015
a
so so hang
(100-1):1+1=100(so hang)
tong bang
(100+1)x100:2=5050
a) \(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\)
\(\Rightarrow2S=1+\frac{1}{2}+...+\frac{1}{2^{2013}}\)
\(\Rightarrow2S-S=1-\frac{1}{2^{2014}}\)
b) Ta có : \(S=1-\frac{1}{2^{2013}}< 1\left(ĐPCM\right)\)