cho a= 2/5x7 + 2/8x10 + ... + 2/2015 x 2017 so sanh a voi 13%
VD
Những câu hỏi liên quan
Tinh A= 2014/2015+2015/2016+2016/2017+2017/2014 hay so sanh A voi 4
=(2014/2014)+(2015+2015)+(2016/2016)+(2017+2017)
=1+1+1+1
=4
vậy A=4 (4=4)
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a. So sanh 2 phan so:A= 2015/2016+2016/2017+2017/2018 va B = 2015+2016+2017/2016+2017+2018
b.1/2.4+1/4.6+........+1/(2x-2).2x = 1/8
c.Cho A = 1/4+1/9+1/16+...+1/81+1/100 . Chung minh rang : A > 65/132
d.Cho B = 12/(2 . 4 ) ^ 2 + 20/ (4 . 6) ^2 + ...........+ 388/ ( 96 . 98 ) ^ 2 + 396/ ( 98 . 100 ) ^2 .Hay so sanh B voi 1 /4
Cho tong T=2/2^1+3/2^2+4/2^3 +...+2016/2^2015+2017/2^2016.So sanh T voi 3
Cho tong T=2/2^1+3/2^2+4/2^3 +...+2016/2^2015+2017/2^2016.So sanh T voi 3
ai trả lời được mình tik 3 nick luôn
Ta có :
\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\)
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\(T=1+\frac{3}{1.2^2}+\frac{4}{2.2^2}+\frac{5}{2^2.2^2}+...+\frac{2016}{2^{2013}.2^2}+\frac{2017}{2^{1014}.2^2}\)
\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{4}+\frac{6}{8}+...+\frac{2016}{x}+\frac{2017}{x}\right)\)
\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2013}}+\frac{2017}{2^{2014}}\right)\)
Đến chỗ này chịu!
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Ta có
\(T=1+\frac{3}{1\cdot2^2}+\frac{4}{2\cdot2^2}+...+\frac{2017}{2^2\cdot2^{2014}}\)
\(T=1+\frac{1}{2^2}\cdot\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\right)\)
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cho a = 1/2*2+1/3*3+1/4*4+....+1/2017*2017
so sanh a voi 1
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{2017.2017}\)
Ta có :
\(\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3.3}< \frac{1}{2.3}\)
\(\frac{1}{4.4}< \frac{1}{3.4}\)
........
\(\frac{1}{2017.2017}< \frac{1}{2016.2017}\)
=> \(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{2017.2017}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2016.2017}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}< 1\)
=> A < 1
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\(a=\frac{1}{2.2}+\frac{1}{3.3}+........+\frac{1}{2017.2017}\)
\(a< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2016.2017}\)
\(a< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2016}-\frac{1}{2017}\)
\(a< 1-\frac{1}{2017}\)
Do \(a< 1-\frac{1}{2017}\)
\(\Rightarrow a< 1\)
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so sanh A voi 1 biet A= 2^2019-(2^2018+2^2017+...+2^1+2^0)
\(A=2^{2019}-\left(2^{2018}+2^{2017}+2^{2016}+.....+2^1+2^0\right)\)
Đặt: \(B=2^{2018}+2^{2017}+2^{2016}+....+2^1+2^0\)
\(\Rightarrow2B=\left(2^{2018}+2^{2017}+2^{2016}+...+2^1+2^0\right)\)
\(\Rightarrow2B-B=\left(2^{2019}+2^{2018}+2^{2017}+...+2^2+2\right)-\left(2^{2018}+2^{2017}+2^{2016}+...+2^1+2^0\right)\)
\(\Rightarrow B=2^{2019}-1\)
\(\Rightarrow A=2^{2019}-\left(2^{2018}+2^{2017}+2^{2016}+.....+2^1+2^0\right)\)
\(=2^{2019}-\left(2^{2019}-1\right)=2^{2019}+2^{2019}+1>1\)
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Mình nhầm ạ ~
\(2^{2019}-\left(2^{2019}-1\right)=2^{2019}-2^{2019}+1=1.\)
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A=(1+1/2015)×(1+1/2015^2)+...+(1+1/2015^2016)
So sanh voi B=2015^2-1/2014^2-1
so sanh
A = 2016^2 va B = 2017 . 2015
so sanh:
a. 2^70 va 3^51
b. 2015/2017 va 2017/2018
351>350=925>825=275>270
Vì 2017<2018 nên\(\frac{1}{2017}\)>\(\frac{1}{2018}\)
⇒\(\frac{2}{2017}\)>\(\frac{1}{2018}\)
⇒\(\frac{2015}{2017}\)=1-\(\frac{2}{2017}\)<1-\(\frac{1}{2018}\)=\(\frac{2017}{2018}\)
Vậy, \(\frac{2015}{2017}\)< \(\frac{2017}{2018}\)