CJ
0
K
Khách
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HV
0
NC
26 tháng 8 2020
Bài làm:
Ta có: \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2017}}\)
=> \(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2016}}\)
=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2016}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2017}}\right)\)
<=> \(2B=1-\frac{1}{3^{2017}}\)
=> \(B=\frac{1}{2}-\frac{1}{3^{2017}.2}< \frac{1}{2}\)
=> \(B< \frac{1}{2}\)
PN
2
24 tháng 11 2017
Ta có :
\(\frac{a1}{a2}=\frac{a2}{a3}=\frac{a3}{a4}=...=\frac{a2016}{a2017}=\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\)
vì \(\frac{a1}{a2}=\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\)
\(\frac{a2}{a3}=\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\)
...
\(\frac{a2016}{a2017}=\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\)
\(\Rightarrow\frac{a1}{a2}.\frac{a2}{a3}.\frac{a3}{a4}...\frac{a2016}{a2017}=\frac{\left(a1+a2+a3+...+a2016\right)^{2016}}{\left(a2+a3+a4+...+a2017\right)^{2016}}\)
\(\Rightarrow\frac{a1}{a2017}=\left(\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\right)^{2016}\)
24 tháng 11 2017
Ta có a1/a2=a2/a3=a3/a4=...=a2016/a2017
=> a1/a2=(a1+a2+a3+...+a2016)
/(a2+a3+a4+...+a2017)
=> a12016/a22016 =(a1+a2+a3+...+a2016)2016 /(a2+a3+a4+...+a2017)2016 (1)
Ta lại có a1/a2=a2/a3=a3/a4=...=a2016/a2017
=> a12016/a22016= a1/a2.a2/a3.a3/a4. ... .a2016/a2017=a1/a2017 (2)
Từ (1) và (2) => đpcm
N
1
KN
16 tháng 4 2019
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 \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}< \frac{2016}{2017}\left(đpcm\right)\)
DN
0
CT
1
12 tháng 12 2017
Ta có :
\(S=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2016}+\left(\frac{1}{2}\right)^{2017}\)
\(2S=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2015}+\left(\frac{1}{2}\right)^{2016}\)
\(2S-S=\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2015}+\left(\frac{1}{2}\right)^{2016}\right]-\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2016}+\left(\frac{1}{2}\right)^{2017}\right]\)
\(S=1-\left(\frac{1}{2}\right)^{2017}< 1\)
9 tháng 12 2019
Đặt A = \(\frac{1.3+2}{2^2}+\frac{2.4+2}{3^2}+\frac{3.5+2}{4^2}+...+\frac{2010.2012+2}{2011^2}+\frac{2015.2017+2}{2016^2}\)
\(=\frac{\left(2-1\right)\left(2+1\right)+2}{2^2}+\frac{\left(3-1\right)\left(3+1\right)}{3^2}+...+\frac{\left(2016-1\right)\left(2016+1\right)+2}{2016^2}\)
\(=\frac{2^2-1+2}{2^2}+\frac{3^2-1+2}{3^2}+....+\frac{2016^2-1+2}{2016^2}\)
\(=\frac{2^2+1}{2^2}+\frac{3^2+1}{3^2}+...+\frac{2016^2+1}{2016^2}\)
\(=\left(1+1+...+1\right)+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}\right)\)(2015 hạng tử 1)
\(=2015+\left(\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{2016.2016}\right)\)
\(< 2015+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\right)\)
\(=2015+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\right)=2015+\left(1-\frac{1}{2016}\right)\)
= 2015 + 1 + 1/2016
= 2016 + 1/2016 < 2017
=> A < 2017 (ĐPCM)