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)

1 < S < 2

=> S ko phải là số tự nhiên

1< S< 2

=> S không phải số tự nhiên

1 < S < 2

\(\Rightarrow\) S ko fai là số tự nhiên

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

thanks, love you 3000!!!!!!!!!!!!!!!!

\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-2.\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)

\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)

\(S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=P-1\)

\(\Rightarrow\left(S-P\right)^{2018}=\left(P-1-P\right)^{2018}=\left(-1\right)^{2018}=1\)

 help me

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

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) \(S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(\Rightarrow5S=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(5S-S=1-\frac{1}{5^{100}}\)

\(4S=1-\frac{1}{5^{100}}\)

\(S=\frac{1-\frac{1}{5^{100}}}{4}\)

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d) \(S=\frac{1!}{3!}+\frac{2!}{4!}+\frac{3!}{5!}+...+\frac{2018!}{2020!}\)

\(S=\frac{1}{1.2.3}+\frac{1.2}{1.2.3.4}+\frac{1.2.3}{1.2.3.4.5}+...+\frac{1.2.3...2018}{1.2.3...2020}\)

\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2019.2020}\)

\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(S=\frac{1}{2}-\frac{1}{2020}\)

\(S=\frac{1009}{2020}\)