NT
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
PN
1
CT
1
12 tháng 12 2017
Ta có :
\(S=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2016}+\left(\frac{1}{2}\right)^{2017}\)
\(2S=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2015}+\left(\frac{1}{2}\right)^{2016}\)
\(2S-S=\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2015}+\left(\frac{1}{2}\right)^{2016}\right]-\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2016}+\left(\frac{1}{2}\right)^{2017}\right]\)
\(S=1-\left(\frac{1}{2}\right)^{2017}< 1\)
HN
0
K
0
NT
4
13 tháng 7 2018
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2018}}\right)\)
\(A=1-\frac{1}{2^{2018}}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
hok tốt .
TT
0
BN
0
DA
2
MT
10 tháng 5 2016
\(\frac{1}{1+2+3+...+n}=\frac{1}{\frac{\left(1+n\right).n}{2}}=\frac{2}{\left(1+n\right).n}=2.\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
áp dụng vào mà làm
HP
10 tháng 5 2016
Ta có công thức: \(1+2+3+....+n=\frac{n.\left(n+1\right)}{2}\)
Áp dụng vào tình tổng S:
\(S=1+\frac{1}{\frac{2.\left(2+1\right)}{2}}+\frac{1}{\frac{3.\left(3+1\right)}{2}}+.....+\frac{1}{\frac{n.\left(n+1\right)}{2}}\)
\(S=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+.....+\frac{1}{\frac{n.\left(n+1\right)}{2}}\)
\(S=1+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{n\left(n+1\right)}\)
Đặt \(A=\frac{2}{2.3}+\frac{2}{3.4}+.....+\frac{2}{n\left(n+1\right)}\) ,ta có:
\(\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n\left(n+1\right)}\)
\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n}-\frac{1}{n+1}=\frac{1}{2}-\frac{1}{n+1}=\frac{n+1-2}{2\left(n+1\right)}=\frac{n-1}{2n+2}\)
=>\(A=\frac{n-1}{2n+2}.2=\frac{2\left(n-1\right)}{2n+2}=\frac{2n-2}{2n+2}=\frac{2n+2-4}{2n+2}=1-\frac{4}{2n+2}<1\)
=>A < 1
Mà S=1+A
=>S < 2 (đpcm)
TT
1
28 tháng 12 2018
\(S=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+.......+\left(\frac{1}{2}\right)^{2017}\)
\(\Rightarrow2S=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+......+\left(\frac{1}{2}\right)^{2016}\)
\(\Rightarrow2S-S=1-\left(\frac{1}{2}\right)^{2017}\)
\(\Rightarrow S=1-\left(\frac{1}{2}\right)^{2017}< 1\left(đpcm\right)\)
\(2S=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{2017}\)
\(S=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2018}\)
\(\Rightarrow S=2S-S=1-\left(\frac{1}{2}\right)^{2018}\)
\(\Rightarrow S< 1\)( đpcm )
\(S=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2018}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\)
\(2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)
\(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\right)\)
\(S=1-\frac{1}{2^{2018}}< 1\) ( đpcm )
Chúc bạn học tốt ~