\(P=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2008}{3^{2008}}+\frac{2009}{3^{2009}}\)

\(\Rightarrow3P=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2009}{3^{2008}}\)

\(\Rightarrow2P=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2008}}-\frac{2009}{3^{2009}}=A-\frac{2009}{3^{2009}}\)

\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

\(\Rightarrow3A=3+1+\frac{1}{3}+...+\frac{1}{3^{2007}}\)

\(\Rightarrow2A=3-\frac{1}{3^{2008}}< 3\Rightarrow A< \frac{3}{2}\)

\(\Rightarrow2P=A-\frac{2009}{2^{2009}}< A< \frac{3}{2}\Rightarrow P< \frac{3}{4}\)

S=1+3+3^2+3^3+3^4+...+3^2009

=(1+3)+(3^2+3^3)+...+(3^2008+3^2009)

=4+3^2(1+3)+...+3^2008(1+3)

=4(1+3^2+...+3^2008) chia hết cho 4

Đầu tiên ta chứng minh \(\frac{1}{n.n}< \frac{1}{\left(n-1\right).\left(n+1\right)}\)(n thuộc N*)

Ta có: \(\frac{1}{\left(n-1\right).\left(n+1\right)}=\frac{1}{\left(n-1\right).n+\left(n-1\right)}=\frac{1}{n.n-n+n-1}=\frac{1}{n.n-1}>\frac{1}{n.n}\)

\(S=\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2009^3}< \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2008.2009.2010}\)

\(S< \frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2008.2009.2010}\right)\)

                                                                   \(S< \frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2008.2009}-\frac{1}{2009.2010}\right)\)

\(S< \frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2009.2010}\right)\)

\(S< \frac{1}{2}.\frac{1}{2}=\frac{1}{4}\)

=> S < 1/4 (đpcm)

Ủng hộ mk nha ^_-

Dễ quá, thực hiện qui tắc bỏ dấu ngoặc được:

 \(2009+2009^2+....+2009^{2009}-1-2009-...-2009^{2008}\)

\(=-1+\left(2009-2009\right)+\left(2009^2-2009^2\right)+...+\left(2009^{2008}-2009^{2008}\right)+2009^{2008}\)

\(=2009^{2008}-1\)

\(=\left(2009-1\right)\left(2009^{2007}+2009^{2008}+...+2009+1\right)\)

\(=2008\left(2009^{2007}+2009^{2008}+...+2009+1\right)\) chia hết cho 2008

=> ĐPCM

Chứng Minh Rằng: (2009+2009²+2009³+2009⁴+...+2009²⁰⁰⁹)-(1+2009+2009²+2009³+...+2009²⁰⁰⁸) chia hết cho 2008.

Đặt A=2009+2009²+2009³+2009⁴+...+2009²⁰⁰⁹, B=1+2009+2009²+2009³+2009⁴+...+2009²⁰⁰⁸

Ta có:

+)A=2009+2009²+2009³+2009⁴+...+2009²⁰⁰⁹

  2009A= 2009²+2009³+2009⁴+...+2009²⁰¹⁰

   2009A-A=(2009²+2009³+2009⁴+...+2009²⁰¹⁰)-(2009+2009²+2009³+2009⁴+...+2009²⁰⁰⁹)

  2008A=2009²⁰¹⁰- 2009

=> A=(2009²⁰¹⁰- 2009)/2008 

=> A chia hết cho 2008.

B=1+2009+2009²+2009³+2009⁴+...+2009²⁰⁰⁸

2009B=2009+2009²+2009³+2009⁴+...+2009²⁰¹⁰

2009B-B=(2009+2009²+2009³+2009⁴+...+2009²⁰¹⁰)-(1+2009+2009²+2009³+2009⁴+...+2009²⁰⁰⁹)

2008B=2009²⁰¹⁰-1

=>B=(2009²⁰¹⁰-1)/2008

=>B chia hết cho 2008

=> A-B chia hết cho 2008.

=> ĐPCM

S = 1 + 3 + 3^2 + 3^3 + 3^4 + .... + 3^2009

S = 3^0 + 3^1 + 3^2 + 3^3 + 3^4 + .... + 3^2009

Từ 0 -> 2009 có tất cả số  số hạng là : 

( 2009 - 0 ) : 1 + 1 = 2010 ( số )

=> có : 2010 : 2 = 1005 cặp

=> S = ( 3^0 + 3^1 ) + ( 3^2 + 3^3 ) + ( 3^4 + 3^5 ) + .... + ( 3^2008 + 3^2009 )

=> S = ( 1 + 3 ) + ( 9 + 27 ) + ( 81 + 243 ) + ....

=> S = 4 + 36 + 324 + ....

Ta thấy 4 ; 36 ; 324 đều chia hết cho 4 => ( 3^0 + 3^1 ) + ( 3^2 + 3^3 ) + ( 3^4 + 3^5 ) chia hết cho 4

=> 3^2008 + 3^2009

=>  ( 3^0 + 3^1 ) + ( 3^2 + 3^3 ) + ( 3^4 + 3^5 )  + .... + ( 3^2008 + 3^2009 ) chia hết cho 4

=> S chia hết cho 4

Vậy ... 

( MK làm theo suy nghĩ có gì trình bày sai or gì đó bạn có thể sửa lại !! ^^

chia hết cho mấy thế bạn