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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:3B\(\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+...+\frac{1}{3}^{2003}+\frac{1}{3}^{2004}\)
B=\(\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+..+\frac{1}{3}^{2003}+\frac{1}{3}^{2004}+\frac{1}{3}^{2005}\)
\(\Rightarrow\)2B=1-\(\frac{1}{3}^{2005}\)
\(\Rightarrow\)B=\(\frac{1-\frac{1}{3}^{2005}}{2}\)
\(\Rightarrow\)B=\(\frac{1-\frac{1}{3}^{2005}}{2}<\frac{1}{2}\)
\(\Rightarrow\)B<\(\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Leftrightarrow2B=3\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)
\(\Leftrightarrow2B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Leftrightarrow2B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)
\(\Leftrightarrow B=1-\frac{1}{3^{2005}}\)
\(\Leftrightarrow B=1-\frac{1}{3^{2005}}< \frac{1}{2}\)
Vậy \(B< \frac{1}{2}\) (Đpcm)
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+..+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\\ \)
\(C=3B=1+\dfrac{1}{3}+..+\dfrac{1}{3^{2004}}\)
\(C-B=1-\dfrac{1}{3^{3005}}\)
\(B=\dfrac{1}{2}-\dfrac{1}{2.3^{2005}}< \dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A xp=x+x2+x^3+x^4+..................+x^2016
=>xp-p= x^2016-1 ban nhe
B ap dung dau hieu chia het cho 3 la tong chu so chia het cho 3
![](https://rs.olm.vn/images/avt/0.png?1311)
B1 a, a^3 - a = a.(a^2-1) = (a-1).a.(a+1) chia hết cho 3
b, a^7-a = a.(a^6-1) = a.(a^3-1).(a^3+1)
Ta thấy số lập phương khi chia 7 dư 0 hoặc 1 hoặc 6
+Nếu a^3 chia hết cho 7 => a^7-a chia hết cho 7
+Nếu a^3 chia 7 dư 1 thì a^3-1 chia hết cho 7 => a^7-a chia hết cho 7
+Nếu a^3 chia 7 dư 6 => a^3+1 chia hết cho 7 => a^7-a chia hết cho 7
Vậy a^7-a chia hết cho 7
b, a^7-a=a(a^6-1)
=a(a^3+1)(a^3-1)
=a(a+1)(a^2-a+1)(a-1)(a^2+a+1)
=a(a-1)(a+1)(a^2-a+1)(a^2+a+1)
=a(a-1) (a+1) (a^2-a+1-7) (a^2+a+1)
+7a (a-1) (a+1) (a^2+a-1)
=a (a-1) (a+1) (a^2-a-6) (a^2+a+1-7)
+7a (a-1) (a+1) (a^2+a-1)
+7a (a-1) (a+1) (a^2-a-6)
có: 7a(a-1) (a+1) (a^2+a-1)+7a (a-1) (a+1) (a^2-a-6) chia hết cho 7 (cùng có nhân tử 7)
ta cần chứng minh: a(a-1) (a+1) (a^2-a-6) (a^2+a+1-7) chia hết cho 7
thật vậy: a(a-1) (a+1) (a^2-a-6) (a^2+a+1-7)
=a(a-1) (a+1) [(a+2)(a-3)] [(a-2)(a+3)]
=(a-3) (a-2) (a-1) a (a+1) (a+2) (a+3) là tích của 7 số nguyên liên tiếp nên chia hết cho 7.
trong 7 số tự nhiên liên tiếp có 1 số chia hết cho 7,1 số dư 1,1 số dư 2,....và 1 số dư 6 khi chia cho 7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}+\frac{1}{3^{2015}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}+\frac{1}{3^{2015}}\right)\)
\(2B=1-\frac{1}{3^{2015}}\)
\(B=\frac{1-\frac{1}{3^{2015}}}{2}\)
Mà \(1-\frac{1}{3^{2015}}
Câu của đặng phương thảo sai rồi ở 3b-b thì là 3^2005 chứ không phải là 3^ 2015
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow3B=3+\frac{1}{3^1}+\frac{1}{3^2}+....+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=\left(3+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=3-\frac{1}{3^{2005}}\Rightarrow B=\left(3-\frac{1}{3^{2005}}\right):2\)
\(\Rightarrow\left(3-\frac{1}{3^{2005}}\right):2<\frac{1}{2}\Rightarrow B<\frac{1}{2}\)
3B=1+1/3+1/32+...+1/32004
3B-B=1-1/32005
2B=1-1/32005
B=1/2-1/(32005.2)
Vậy B <1/2
a)ta có 3B=1+1/3+1/3^2+........+1/3^2003+1/3^2004
B= 1/3+1/3^2+........+1/3^2003+1/3^2004+1/3^2005
suy ra 2B=1-1/3^2005
suy ra B=\(\frac{1-\frac{1}{3}^{2005}}{2}\)
suy ra B=1/2-1/3^2005/2 bé hơn 1/2
từ đấy suy ra B bé hơn 1/2