ĐẶT A=\(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

  \(\frac{1}{3}A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}\)

\(\frac{1}{3}A-A=\frac{1}{3^{2006}}-\frac{1}{3^0}\)

\(\frac{-2}{3}A=\frac{1}{3^{2006}}-\frac{1}{3^0}\)

\(A=\frac{\frac{1}{3^{2006}}-1}{\frac{-2}{3}}\)

D=1+1/3^1+1/3^2+...+...+1/3^2005

3D=3+1/3^2+1/3^1+...+...+1/3^2006

3D-D=(3+1+1/3^1+...+...+1/3^2004)-(1+1/3+1/3^2+...+...+1/3^2005)

2D=3-1/3^2005

D=(3-1/3^2005):2

CÓ PHẢI TÍNH RA KẾT QUẢ KHÔNG ?NẾU PHẢI TÍNH THÌ CẬU TÍNH GIÙM MÌNH NHÉ!

\(A=\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

\(\Rightarrow3A=1+\frac{1}{3^0}+\frac{1}{3^1}+...+\frac{1}{3^{2004}}\)

\(\Rightarrow2A=1-\frac{1}{3^{2005}}\)

\(\Rightarrow A=\frac{3^{2005}-1}{3^{2005}.2}\)

cảm ơn thành đạt 

\(G=\frac{1}{3^0}+\frac{1}{3^1}+...+\frac{1}{3^{2005}}\)\(\Rightarrow3G=3+\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)

\(\Rightarrow3G-G=2G=3-\frac{1}{3^{2005}}\)\(\Rightarrow G=\frac{3-\frac{1}{3^{2005}}}{2}\)

\(Y=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)\(\Rightarrow2Y=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

\(\Rightarrow2Y-Y=2-\frac{1}{2^{2012}}\) \(\Rightarrow Y=2-\frac{1}{2^{2012}}\)

Cảm ơn bạn nhiều lắm 

http://olm.vn/thanhvien/hai2804 ngu nhu bo tot

tổng của nó là 2006

Ta có :

\(Y=\dfrac{1}{3^0}+\dfrac{1}{3^1}+\dfrac{1}{3^2}+................+\dfrac{1}{3^{2005}}\)

\(\Rightarrow3Y=2+\dfrac{1}{3^0}+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...........+\dfrac{1}{3^{2004}}\)

\(\Rightarrow3Y-Y=\left(2+\dfrac{1}{3^0}+\dfrac{1}{3^1}+.............+\dfrac{1}{3^{2004}}\right)-\left(\dfrac{1}{3^0}+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...........+\dfrac{1}{3^{2005}}\right)\)\(\Rightarrow2Y=2-\dfrac{1}{3^{2005}}\)

\(\Rightarrow Y=\dfrac{2-\dfrac{1}{3^{2005}}}{2}\)

a) \(1+2+3+4+...+n\)

\(=\left(n+1\right)\left[\left(n-1\right):1+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right):2\)

\(=n\left(n+1\right):2\)

\(=\dfrac{n\left(n+1\right)}{2}\)

b) \(2+4+6+..+2n\)

\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)

\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right)\)

\(=n\left(n+1\right)\)

c) \(1+3+5+...+\left(2n+1\right)\)

\(=\left[\left(2n+1\right)+1\right]\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}:2\)

\(=\left(2n+1+1\right)\left[\left(2n-1-1\right):2+1\right]:2\)

\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)

\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right)\)

\(=n\left(n+1\right)\)

d) \(1+4+7+10+...+2005\)

\(=\left(2005+1\right)\left[\left(2005-1\right):3+1\right]:2\)

\(=2006\cdot\left(2004:3+1\right):2\)

\(=2006\cdot\left(668+1\right):2\)

\(=1003\cdot669\)

\(=671007\)

e) \(2+5+8+...+2006\)

\(=\left(2006+2\right)\left[\left(2006-2\right):3+1\right]:2\)

\(=2008\cdot\left(2004:3+1\right):2\)

\(=1004\cdot\left(668+1\right)\)

\(=1004\cdot669\)

\(=671676\)

g) \(1+5+9+...+2001\)

\(=\left(2001+1\right)\left[\left(2001-1\right):4+1\right]:2\)

\(=2002\cdot\left(2000:4+1\right):2\)

\(=1001\cdot\left(500+1\right)\)

\(=1001\cdot501\)

\(=501501\)

\(3S=3+\frac{1}{3}+...+\frac{1}{3^{2004}}\)

\(3S-S=\left(3+\frac{1}{3}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)

\(2S=3-\frac{1}{3^{2005}}\)

\(2S=\frac{3^{2006-1}}{3^{2005}}\)

\(S=\frac{3^{2006}-1}{3^{2005}.2}\)

S = 1/3 + 1/3² + 1/3³ + ... + 1/3²⁰⁰⁵

=> 3S = 1 + 1/3 + 1/3² + ... + 1/3²⁰⁰⁴

=> 3S - S = 1 + 1/3 + 1/3² + ... + 1/3²⁰⁰⁴ - (1/3 + 1/3² + 1/3³ + ... + 1/3²⁰⁰⁵)

=> 2S = 1 + 1/3 + 1/3² + ... + 1/3²⁰⁰⁴ - 1/3 - 1/3² - 1/3³ - ... - 1/3²⁰⁰⁵

=> 2S = 1 - 1/3²⁰⁰⁵

=> S = \(\frac{\frac{1}{3^{2005}}}{2}\)

=> S = 1/3²⁰⁰⁵.2