cho S=\(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2018}{4^{2018}}\)CMR: S<\(\frac{1}{2}\)
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AH
Akai Haruma
Giáo viên
23 tháng 2 2020
Lời giải:
$D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+......+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}$
$4D=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}$
Trừ theo vế:
\(3D=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow 12D=4+1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2017}}-\frac{2019}{4^{2018}}\)
Trừ theo vế:
$9D=4-\frac{2019}{4^{2018}}+\frac{2019}{4^{2019}}-\frac{1}{4^{2018}}$
$=4-\frac{6061}{4^{2019}}< 4$
$\Rightarrow D< \frac{4}{9}<\frac{4}{8}$ hay $D< \frac{1}{2}$ (đpcm)
XO
14 tháng 3 2020
Ta có : \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2018}{3^{2018}}\)(1)
\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{2018}{3^{2019}}\)(2)
Lấy (1) trừ (2) theo vế ta có :
\(A-\frac{1}{3}A=\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2018}{3^{2018}}\right)-\left(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{2018}{3^{2019}}\right)\)
\(\Rightarrow\frac{2}{3}A=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\right)-\frac{2018}{3^{2019}}\)
Đặt B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\)
=> 3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\)
Lấy 3B trừ B theo vế ta có :
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\right)\)
=> 2B = \(1-\frac{1}{3^{2018}}\)
=> \(B=\frac{1}{2}-\frac{1}{3^{2018}.2}\)
Khi đó : \(\frac{2}{3}A=\frac{1}{2}-\frac{1}{3^{2018}.2}-\frac{2018}{3^{2019}}\)
\(A=\left(\frac{1}{2}-\frac{1}{3^{2018}.2}-\frac{2018}{3^{2019}}\right):\frac{2}{3}=\frac{3}{4}-\frac{1}{3^{2017}.4}-\frac{1009}{3^{2018}}=\frac{3}{4}-\left(\frac{1}{3^{2017}.\left(3+1\right)}+\frac{1009}{3^{2018}}\right)\)
\(=\frac{3}{4}-\left(\frac{1}{3^{2018}}+\frac{1}{3^{2017}}-\frac{1009}{3^{2018}}\right)=\frac{3}{4}-\left(\frac{1}{3^{2017}}-\frac{336}{3^{2017}}\right)=\frac{3}{4}+\frac{335}{3^{2017}}\)
Vì A > 0 (1)
Mặt khác\(\frac{335}{3^{2017}}< \frac{335}{1340}< \frac{1}{4}\)
=> \(\frac{335}{3^{2017}}< \frac{1}{4}\Rightarrow\frac{3}{4}+\frac{335}{3^{2017}}< \frac{1}{4}+\frac{3}{4}\Rightarrow A< 1\)(2)
Từ (1) và (2) => 0 < A < 1
=> A không phải là số nguyên
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 4 2019
\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-2.\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=P-1\)
\(\Rightarrow\left(S-P\right)^{2018}=\left(P-1-P\right)^{2018}=\left(-1\right)^{2018}=1\)
KT
6 tháng 1 2019
Đặt: \(A=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2018}}\)
Ta có: \(\frac{1}{\sqrt{k}}=\frac{2}{\sqrt{k}+\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}=2\left(\sqrt{k+1}-\sqrt{k}\right)\) với \(\forall k\inℕ^∗\)
Do đó ta có: \(A>2\left[\left(\sqrt{2019}-\sqrt{2018}\right)+\left(\sqrt{2018}-\sqrt{2017}\right)+...+\left(\sqrt{3}-\sqrt{2}\right)\right]+1\)
\(=2\left(\sqrt{2019}-\sqrt{2}\right)+1=2\sqrt{2019}-2\sqrt{2}+1>2\sqrt{2019}-3+1>2\sqrt{2019}-2\)
\(>2\sqrt{2018}-2=2\left(\sqrt{2018}-1\right)\)
=> đpcm
24 tháng 6 2018
a) 2 +4+6+8+...+2018
= ( 2018+2) x 1009 : 2
= 2020 x 1009 : 2
= 1009 x (2020:2)
= 1009 x 1010
= 1 019 090
b) S = 10 + 102 + 103 + ...+ 10100
=> 10.S = 102 + 103 + 104 +...+ 10101
=> 10.S - S = 10101-10
9.S=10101- 10
\(\Rightarrow S=\frac{10^{101}-10}{9}\)
c) \(S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(\Rightarrow5S=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(5S-S=1-\frac{1}{5^{100}}\)
\(4S=1-\frac{1}{5^{100}}\)
\(S=\frac{1-\frac{1}{5^{100}}}{4}\)
e cx ko nx, e ms hok lp 7 thoy, sang hè ms lp 8! e sr cj nhiều nha!
24 tháng 6 2018
d) \(S=\frac{1!}{3!}+\frac{2!}{4!}+\frac{3!}{5!}+...+\frac{2018!}{2020!}\)
\(S=\frac{1}{1.2.3}+\frac{1.2}{1.2.3.4}+\frac{1.2.3}{1.2.3.4.5}+...+\frac{1.2.3...2018}{1.2.3...2020}\)
\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2019.2020}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(S=\frac{1}{2}-\frac{1}{2020}\)
\(S=\frac{1009}{2020}\)