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Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
Ta có A= 1/2015 + 2/2016 + 3/2017 + ... +2016/4030- 2016
A= 2015-2014/2015 + 2016-2014/2016 +...+4030-2014/4030-2016
A= 2015/2015-2014/2015+ 2016/2016-2014/2016 + ..... +4030/4030-2014/4030 -2016
A= 1-2014/2015 + 1-2014/2016 +....+1-2014/4030 -2016
A= (1+1+1+1+........+1) -(2014/2015+2014/2016+......+2014/4030) -2016
A=2016 - 2014.(1/2015+1/2016+....+1/4030) -2016
A= (2016 - 2016 ) - 2014. ( 1/2015+1/2016+.....+1/4030)
A=-2014.(1/2015+1/2016+....+1/4030)
mà B = 1/2015+1/2016+....+1/4030
nên A : B = -2014
\(3.M=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{38}}\)
=> \(3M-M=2M=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{38}}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^{39}}\)
=> \(2M=1-\frac{1}{3^{39}}\)
=> \(M=\frac{1}{2}\left(1-\frac{1}{3^{39}}\right)\)
do \(1-\frac{1}{3^{39}}< 1\)
=> \(\frac{1}{2}\left(1-\frac{1}{3^{39}}\right)< \frac{1}{2}.1=\frac{1}{2}\)
Vay \(M< \frac{1}{2}\)
Chuc bn hoc tot !
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A< 1-\frac{1}{2017}=\frac{2016}{2017}\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}< \frac{2016}{2017}\left(đpcm\right)\)
vì -1 hơn 1 hai số cho nên;
a) a/b và c/d ^2 =ab/cd hơn kém nhau 2
b) dựa theo tính chất kết hợp (a+b/c+d ) ^3 = a ^3 ...
Bài làm:
Ta có: \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2017}}\)
=> \(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2016}}\)
=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2016}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2017}}\right)\)
<=> \(2B=1-\frac{1}{3^{2017}}\)
=> \(B=\frac{1}{2}-\frac{1}{3^{2017}.2}< \frac{1}{2}\)
=> \(B< \frac{1}{2}\)