tks bn

a,Ta có: A có 2016 số số hạng, ghép A thành 504 nhóm, mỗi nhóm có 4 số hạng như sau :

\(A=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+....+(3^{2013}+3^{2014}+3^{2015}+3^{2016})\)

\(A=3.(1+3+3^2)+3^5.(1+3+3^2)+....+3^{2013}.(1+3+3^2)\)

\(A=3.13+3^5.13+....+3^{2013}.13\)

\(A=13.(3+3^5+...+3^{2013})⋮13\)

\(\Rightarrow A⋮13\)

\(a\)) Ta có :

\(A=3+3^2+3^3+..........+3^{2016}\) (2016 số hạng )

\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+.....+\left(3^{2014}+3^{2015}+3^{2016}\right)\) (672 nhóm )

\(A=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+.......+3^{2015}\left(1+3+3^2\right)\)

\(A=3.13+3^4.13+........+3^{2015}.13\)

\(A=13\left(3+3^4+.......+3^{2016}\right)\)

\(\Rightarrow A\) \(⋮\) \(13\)

\(\Rightarrowđpcm\)

\(b\)) Ta có :

\(A=3+3^2+3^3+..........+3^{2016}\)

\(\Rightarrow3A=3^2+3^3+...............+2^{2016}+3^{2017}\)

\(\Rightarrow3A-A=3^{2017}-3\)

\(\Rightarrow2A=3^{2017}-3\)

\(\Rightarrow2A+3=3^{2017}\)(1)

Theo bài ta có :

\(2A+3=3^{2x}\)(2)

Từ (1) và (2) ta có :

\(3^{2x}=3^{2017}\)

\(\Rightarrow2x=2017\)

\(x=2017:2\)

\(x=1008,5\) ( ko thoả mãn \(x\in N\))

Vậy ko tìm dc giá trị của \(x\) thỏa mãn theo yêu cầu

Ta có: \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)

\(\Rightarrow A>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)

\(\Rightarrow A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(\Rightarrow A>\frac{1}{2}-\frac{1}{11}=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)

- Đến đây bn lấy \(\frac{9}{22}\) so sánh vs \(\frac{65}{132}\) là ra ĐPCM nhé :3

Ta có: 3A=3+\(^{3^2+3^3+3^4+3^5+...+3^{2012}+3^{2013}}\)

\(\Rightarrow\)3A-A=2A=(\(3+3^2+3^3+3^4+...+3^{2013}\)) - (\(1-3^{ }-3^2-3^3-3^4-...-3^{2012}\))

\(\Rightarrow\)2A=\(3^{2013}-1\)\(\Rightarrow\)A=\(\left(3^{2013}-1\right):2\)\(\Rightarrow\)B-A=(\(^{\left(3^{2013}:2\right)-\left(\left(3^{2013}-1\right):2\right)\Rightarrow}\)

A = 1 + 3 + 3² +...+ 3²⁰¹²

3A = 3 + 3² + 3³ +...+ 3²⁰¹³

3A - A = (3 + 3² + 3³ +...+ 3²⁰¹³) - (1 + 3 + 3² +...+ 3²⁰¹²)

2A = 3²⁰¹³ - 1

A = \(\frac{3^{2013}-1}{2}\)

=> B - A = \(\frac{3^{2013}}{2}-\frac{3^{2013}-1}{2}=\frac{3^{2013}-\left(3^{2013}-1\right)}{2}=\frac{3^{2013}-3^{2013}+1}{2}=\frac{1}{2}\)