a, 5M = 5+1+1/5+1/5^2+.....+1/5^2011

4M=5M-M=(5+1+1/5+1/5^2+.....+1/5^2011)-(1+1/5+1/5^2+.....+1/5^2012)

               = 5-1/5^2012

=> M = (5 - 1/5^2012)/4

Tk mk nha

10.

Sửa lại đề :Cho \(P=\dfrac{2009}{2010}+\dfrac{2010}{2011}+\dfrac{2012}{2013}+\dfrac{2013}{2009}\).Chứng tỏ rằng P<5.

\(P=\dfrac{2009}{2010}+\dfrac{2010}{2011}+\dfrac{2012}{2013}+\dfrac{2013}{2009}\)

\(P=\dfrac{2011}{2012}\)

\(\Rightarrow P< 5\)

M = 2012 + 2012² + ... + 2012²⁰¹⁰

= ( 2012 + 2012² ) + ... + ( 2012²⁰⁰⁹ + 2012²⁰¹⁰ )

= 2012( 1 + 2012 ) + ... + 2012²⁰⁰⁹( 1 + 2012 )

= 2012.2013 + ... + 2012²⁰⁰⁹.2013

= 2013( 2012 + ... + 2012²⁰⁰⁹ ) chia hết cho 2013

hay M chia hết cho 2013 ( đpcm )

Ta có: x=2011 \(\Rightarrow\)x+1=2012

\(\Rightarrow A=x^{2011}-\left(x+1\right).x^{2010}\)\(+\left(x+1\right)x^{2009}\)\(-\left(x+1\right)x^{2008}+...\)\(-\left(x+1\right)x^2+\left(x+1\right)x-1\)

=\(x^{2011}\)\(-x^{2011}-x^{2010}+x^{2010}+x^{2009}-x^{2009}-\)...\(-x^2+x^2+x-1\)

= \(x-1=2011-1=2010\)

=

Thay 2012=x+1.

\(A=x^{2011}-\left(x+1\right)x^{2010}+\left(x+1\right)x^{2009}-\left(x+1\right)x^{2008}+...-\left(x+1\right)x^2+\left(x+1\right)x-1\)

\(A=x^{2011}-x^{2011}-x^{2010}+x^{2010}+x^{2009}-...-x^3-x^2+x^2+x-1\)

\(A=x-1=2011-1=2010\)

mình giải ra trc mà, k mik chứ

Lời giải:

$A=1-\frac{1}{2011}+1-\frac{1}{2012}+1+\frac{2}{2010}$

$=3+(\frac{1}{2010}-\frac{1}{2011})+(\frac{1}{2010}-\frac{1}{2012})$

$> 3+0+0+0=3$

Ta có đpcm.