So sánh tổng S=\(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+.....+\frac{n}{2^n}+.....+\frac{2017}{2^{2017}}\) với 2 (n khác 0)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
LV
11 tháng 11 2017
\(S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+.....+\frac{n}{2^n}+......+\frac{2017}{2^{2017}}\)
Với n > 2 thì \(\frac{n}{2^n}=\frac{n+1}{2^{n-1}}-\frac{n+2}{2^n}\)
\(\frac{n+1}{2^{n-1}}=\frac{n+1}{2^n:2}=\frac{n+1}{\frac{2^n}{2}}=\frac{2^{\left(n+1\right)}}{2^n}\)
\(\frac{n+1}{2^{n-1}}-\frac{n+2}{2^n}=\frac{2^{n+2}}{2^n}-\frac{n+2}{2^n}\)
\(=\frac{2^{n+2}-n-2}{2^n}\)
\(=\frac{n}{2^n}\)
\(\Leftrightarrow S=\frac{1}{2}+\left(\frac{2+1}{2^{2-1}}-\frac{2+2}{2^2}\right)+.....+\frac{2016+1}{2^{2015}}-\frac{2018}{2^{2016}}\)
\(=\frac{2017+1}{2^{2016}}-\frac{2019}{2^{2017}}\)
\(S=\frac{1}{2}+\frac{3}{2}-\frac{2019}{2017}\)
\(S=2-\frac{2019}{2017}\)
\(\Leftrightarrow S=2-\frac{2019}{2017}< 2\)
Hay \(S< 2\)
14 tháng 11 2018
\(P=\frac{3}{1!\left(1+2\right)+3!}+\frac{4}{2!\left(1+3\right)+4!}+...+\frac{2017}{2015!\left(1+2016\right)+2017!}\)
\(P=\frac{3}{3\left(1!+2!\right)}+\frac{4}{4\left(2!+3!\right)}+...+\frac{2017}{2017\left(2015!+2016!\right)}\)
\(P=\frac{1}{1!+2!}+\frac{1}{2!+3!}+...+\frac{1}{2015!+2016!}\)
Ta có \(a!>\sqrt{a}\)\(\left(a\inℕ;a>1\right)\) do đó :
\(P>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2015}+\sqrt{2016}}\)
\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\)
\(\frac{\sqrt{2016}-\sqrt{2015}}{\left(\sqrt{2016}+\sqrt{2015}\right)\left(\sqrt{2016}-\sqrt{2015}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2016}\)
\(-\sqrt{2015}=\sqrt{2016}-1=\frac{1}{2}+\left(\sqrt{2016}-\frac{3}{2}\right)=\frac{1}{2}+\left(\sqrt{2016}-\sqrt{\frac{9}{4}}\right)>\frac{1}{2}\)
Vậy \(P>\frac{1}{2}\)
Chúc bạn học tốt ~
PS : tự nghĩ bừa thui nhé :))
T1
1
BT
22 tháng 3 2018
\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\) => \(\frac{T}{2}=\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\)
=> \(T-\frac{T}{2}=\left(\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\right)-\left(\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\right)\)
<=> \(\frac{T}{2}=\frac{2}{2^1}+\left(\frac{3}{2^2}-\frac{2}{2^2}\right)+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+...+\left(\frac{2017}{2^{2016}}-\frac{2016}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)
<=> \(\frac{T}{2}=1+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)
Đặt: \(M=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}=>2M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\)
=> \(2M-M=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
=> \(M=\frac{1}{2}-\frac{1}{2^{2016}}< \frac{1}{2}\)
=> \(\frac{T}{2}< 1+\frac{1}{2}-\frac{2017}{2^{2017}}< 1+\frac{1}{2}=\frac{3}{2}\)
=> T < 3
27 tháng 12 2017
A=\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+2}+..........+\frac{2018}{2017^2+2017}\)
>\(\frac{2018}{2017^2+2017}+\frac{2018}{2017^2+2017}+........+\frac{2018}{2017^2+2017}\)
\(=\frac{2018}{2017^2+2017}.2017=\frac{2018.2017}{2017\left(2017+1\right)}=1\) (1)
Lại có:A<\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+1}+.........+\frac{2018}{2017^2+1}\)
\(=\frac{2018}{2017^2+1}.2017=\frac{2018.2017}{2017^2+1}=\frac{2017.\left(2017+1\right)}{2017^2+1}\)
\(=\frac{2017^2+2017}{2017^2+1}=\frac{2017^2+1+2016}{2017^2+1}=1+\frac{2016}{2017^2+1}< 2\) (2)
Từ (1) và (2) suy ra:1 < A < 2
Vậy A không phải là số nguyên
