\(S=2^0+2^1+2^2+...+2^7\)

\(\Rightarrow S=\left(2^0+2^1\right)+2^2\left(2^0+2^1\right)+...+2^6\left(2^0+2^1\right)\)

\(\Rightarrow S=3+2^2.3+...+2^6.3\)

\(\Rightarrow S=3\left(1+2^2+...+2^6\right)⋮3\)

\(\Rightarrow dpcm\)

\(S=\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+...+\frac{5}{49}\)

\(S>5\left(\frac{1}{49}+\frac{1}{49}+...+\frac{1}{49}\right)\)(30 số hạng \(\frac{1}{49}\))

\(\Leftrightarrow S>5.\frac{30}{49}\)

\(\Leftrightarrow S>\frac{150}{49}=3\frac{3}{49}\)

\(\Rightarrow S>3\)

\(\Rightarrow S>\frac{3}{49}\)

Vậy \(3< S\)  (1)

Ta lại có: \(S< 5.\left(\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)\)(30 số hạng)

\(S< \frac{30}{20}.5=\frac{150}{20}=\frac{15}{2}=7\frac{1}{2}\)

\(\Rightarrow S< 7< 8\)

\(\Rightarrow S< \frac{1}{2}\)

Vậy \(S< 8\) (2)

Từ (1) và (2) ta có đpcm

thank you very much

khi minh vua dang

\(S=1-2+2^2-2^3+...+2^{2012}-2^{2013}\)

\(\Rightarrow2S=2-2^2+2^3-2^4+...+2^{2013}-2^{2014}\)

\(\Rightarrow2S+S=2-2^2+2^3-...-2^{2014}+1-2^2-2^3+...-2^{2013}\)

\(\Rightarrow3S=1-2^{2014}\)\(\Rightarrow3S-2^{2014}=1-2^{2015}\)

ta có \(\frac{1}{20}>\frac{1}{27};\frac{1}{21}>\frac{1}{27}...;\frac{1}{26}>\frac{1}{27}\)

=> \(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{27}>\frac{7}{27}+\frac{1}{27}=\frac{8}{27}\)(ĐPcm)

Ta có : \(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{27}\)(8 số hạng)

\(>\frac{1}{27}+\frac{1}{27}+\frac{1}{27}+...+\frac{1}{27}\)(8 số hạng)

\(=\frac{1}{27}\times8\)

\(=\frac{8}{27}\)

\(\Rightarrow\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{27}>\frac{8}{27}\left(đpcm\right)\)

Vì \(\frac{1}{20}< \frac{1}{27};\frac{1}{21}< \frac{1}{27};...;\frac{1}{26}< \frac{1}{27}\)

\(\frac{\Rightarrow1}{20}+\frac{1}{21}+...+\frac{1}{27}>\frac{7}{27}+\frac{1}{27}=\frac{8}{27}\left(ĐPCM\right)\)

Đặt  \(B=\frac{1}{20}+\frac{1}{200}+\frac{1}{200}+....+\frac{1}{200}< C=\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+....+\frac{1}{200}\)

Số các phân số \(\frac{1}{200}\)có trong \(B\)là :

 ( 200 - 21 ) :1 + 1 = 180 ( phân số )

Nên \(B=\frac{1}{20}+180.\frac{1}{200}=\frac{1}{20}+\frac{9}{10}>\frac{9}{10}\)

Do đó , \(C>B>\frac{9}{10}\)nên \(C>\frac{9}{10}\)

Vậy \(C>\frac{9}{10}\left(ĐPCM\right)\)

a) P = 1 + 3 + 3² + ... + 3¹⁰¹

= (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3⁹⁹ + 3¹⁰⁰ + 3¹⁰¹)

= 13 + 3³.(1 + 3 + 3²) + ... + 3⁹⁹.(1 + 3 + 3²)

= 13 + 3³.13 + ... + 3⁹⁹.13

= 13.(1 + 3³ + ... + 3⁹⁹) ⋮ 13

Vậy P ⋮ 13

b) B = 1 + 2² + 2⁴ + ... + 2²⁰²⁰

= (1 + 2² + 2⁴) + (2⁶ + 2⁸ + 2¹⁰) + ... + (2²⁰¹⁶ + 2²⁰¹⁸ + 2²⁰²⁰)

= 21 + 2⁶.(1 + 2² + 2⁴) + ... + 2²⁰¹⁶.(1 + 2² + 2⁴)

= 21 + 2⁶.21 + ... + 2²⁰¹⁶.21

= 21.(1 + 2⁶ + ... + 2²⁰¹⁶) ⋮ 21

Vậy B ⋮ 21

c) A = 2 + 2² + 2³ + ... + 2²⁰

= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2¹⁷ + 2¹⁸ + 2¹⁹ + 2²⁰)

= 30 + 2⁴.(2 + 2² + 2³ + 2⁴) + ... + 2¹⁶.(2 + 2² + 2³ + 2⁴)

= 30 + 2⁴.30 + ... + 2¹⁶.30

= 30.(1 + 2⁴ + ... + 2¹⁶)

= 5.6.(1 + 2⁴ + ... + 2¹⁶) ⋮ 5

Vậy A ⋮ 5

d) A = 1 + 4 + 4² + ... + 4⁹⁸

= (1 + 4 + 4²) + (4³ + 4⁴ + 4⁵) + ... + (4⁹⁷ + 4⁹⁸ + 4⁹⁹)

= 21 + 4³.(1 + 4 + 4²) + ... + 4⁹⁷.(1 + 4 + 4²)

= 21 + 4³.21 + ... + 4⁹⁷.21

= 21.(1 + 4³ + ... + 4⁹⁷) ⋮ 21

Vậy A ⋮ 21

e) A = 11⁹ + 11⁸ + 11⁷ + ... + 11 + 1

= (11⁹ + 11⁸ + 11⁷ + 11⁶ + 11⁵) + (11⁴ + 11³ + 11² + 11 + 1)

= 11⁵.(11⁴ + 11³ + 11² + 11 + 1) + 16105

= 11⁵.16105 + 16105

= 16105.(11⁵ + 1)

= 5.3221.(11⁵ + 1) ⋮ 5

Vậy A ⋮ 5