Cho M=1+1/2+1/2^2+...+1/2^2015+1/2^2016. so sanh M voi 2
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\(M=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2016}}\)
\(\Rightarrow\)\(2M=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2015}}\)
\(\Rightarrow\)\(2M-M=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
\(\Rightarrow\)\(M=2-\frac{1}{2^{2016}}< 2\)
Vậy M < 2
Ta có :
\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\)
\(T=1+\frac{3}{1.2^2}+\frac{4}{2.2^2}+\frac{5}{2^2.2^2}+...+\frac{2016}{2^{2013}.2^2}+\frac{2017}{2^{1014}.2^2}\)
\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{4}+\frac{6}{8}+...+\frac{2016}{x}+\frac{2017}{x}\right)\)
\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2013}}+\frac{2017}{2^{2014}}\right)\)
Đến chỗ này chịu!
2A= 2+2^3+2^4+...+2^2015+2^2016
2A-A=2^2016-1
A=(2^2016-1):2
Vì (2^2016-1):2 bé hơn 2^2016 nên A bé hơn 2^2016
\(M=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}\)
\(>1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)
\(=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1+\frac{1}{2}-\frac{1}{11}\)
\(>1+\frac{1}{2}-\frac{1}{6}=\frac{4}{3}\)
M = 1 + 1/2 + 1/22 + ... + 1/22015 + 1/22016
=> 2.M = 2 + 1 + 1/2 + ... + 1/22014 + 1/22015
=> 2.M = 3 + 1/2 +...+ 1/22014 + 1/22015
=> 2.M - M = 3 -1 + 1/22016
=> M = 2 + 1/22016
=> M > 2 ( Do 2 + 1/22016 > 2 )
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