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

Đặt \(A=\frac15+\frac{2}{5^2}+\cdots+\frac{2016}{5^{2016}}\)

=>\(5A=1+\frac25+\cdots+\frac{2016}{5^{2015}}\)

=>\(5A-A=1+\frac25+\cdots+\frac{2016}{5^{2015}}-\frac15-\frac{2}{5^2}-\cdots-\frac{2016}{5^{2016}}\)

=>\(4A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)

Đặt \(B=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}\)

=>\(5B=1+\frac15+\cdots+\frac{1}{5^{2014}}\)

=>\(5B-B=1+\frac15+\cdots+\frac{1}{5^{2014}}-\frac15-\frac{1}{5^2}-\cdots-\frac{1}{5^{2015}}\)

=>\(4B=1-\frac{1}{5^{2015}}=\frac{5^{2015}-1}{5^{2015}}\)

=>\(B=\frac{5^{2015}-1}{4\cdot5^{2015}}\)

TA có: \(4A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)

\(=1+\frac{5^{2015}-1}{4\cdot5^{2015}}-\frac{2016}{5^{2016}}=1+\frac{5^{2016}-5-8064}{4\cdot5^{2016}}=1+\frac14-\frac{8069}{4\cdot5^{2016}}\)

=>\(4A<1+\frac14=\frac54\)

=>\(A<\frac{5}{16}\)

mà \(\frac{5}{16}<\frac{5}{15}=\frac13\)

nên \(A<\frac13\) (1)

Ta có: \(4A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)

=>\(20A=5+1+\frac15+\cdots+\frac{1}{5^{2014}}-\frac{2016}{5^{2015}}\)

=>\(20A-4A=5+1+\frac15+\cdots+\frac{1}{5^{2014}}-\frac{2016}{5^{2015}}-1-\frac15-\frac{1}{5^2}-\cdots-\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)

=>\(16A=5-\frac{2017}{5^{2015}}-\frac{2016}{5^{2016}}>5\)

=>\(A>\frac{5}{16}\)

=>\(A>\frac{4}{16}=\frac14\) (2)

Từ (1),(2) suy ra 1/4<A<1/3

Ta có :

 \(\dfrac{1}{2}>\dfrac{1}{5}\)

\(\dfrac{1}{3}>\dfrac{1}{5}\)

\(\dfrac{1}{4}>\dfrac{1}{5}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{4}{5}\)

\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\)

\(=\dfrac{7}{6}+\dfrac{1}{4}+\dfrac{1}{5}\)

\(=\dfrac{17}{12}+\dfrac{1}{5}\)

\(=\dfrac{97}{60}\)

\(\dfrac{4}{5}=\dfrac{4.12}{5.12}=\dfrac{48}{60}\)

Mà \(\dfrac{97}{60}>\dfrac{48}{60}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{4}{5}\left(đpcm\right)\).

\(1,Y=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ Y=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ Y=13\left(1+3^3+...+3^{96}\right)⋮13\\ 2,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{2019}\right)\\ A=4\left(1+3^2+...+3^{2019}\right)⋮4\\ 3,\Leftrightarrow2\left(x+4\right)=60\Leftrightarrow x+4=30\Leftrightarrow x=36\)

Giúp mình cả bài 4,5 ở dưới được ko?

Sửa đề: \(\frac{2}{5^2}<\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{9^2}<1\)

Ta có: \(\frac{1}{2^2}>\frac{1}{2\cdot3}=\frac12-\frac13\)

\(\frac{1}{3^2}>\frac{1}{3\cdot4}=\frac13-\frac14\)

...

\(\frac{1}{9^2}>\frac{1}{9\cdot10}=\frac19-\frac{1}{10}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{9^2}>\frac12-\frac13+\cdots+\frac19-\frac{1}{10}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{9^2}>\frac12-\frac{1}{10}=\frac25\) (1)

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{9^2}<\frac{1}{8\cdot9}=\frac18-\frac19\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{9^2}<1-\frac12+\frac12-\frac13+\cdots+\frac18-\frac19=1-\frac19<1\) (2)

Từ (1),(2) suy ra \(\frac25<\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{9^2}<1\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}< \frac{1}{2}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}< \frac{1}{2}\)

\(=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)\)

thiếu đề :(

Đặt \(A=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{60^2}\)

\(A< \dfrac{1}{3^2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{59.60}\)

\(A< \dfrac{1}{3^2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\)

\(A< \dfrac{1}{3^2}+\dfrac{1}{3}-\dfrac{1}{60}\)

\(A< \dfrac{4}{9}-\dfrac{1}{60}< \dfrac{4}{9}\) (đpcm)

Thank you !