Nhầm đề hả bạn 

Lời giải:
Xét tử số:
$X=1+2+2^2+2^3+...+2^{2008}$

$2X=2+2^2+2^3+2^4+....+2^{2009}$

$\Rightarrow 2X-X=(2+2^2+2^3+2^4+....+2^{2009})-(1+2+2^2+...+2^{2008})$

$\Rightarrow X=2^{2009}-1$

$\Rightarrow S=\frac{X}{1-2^{2009}}=\frac{2^{2009}-1}{-(2^{2009}-1)}=-1$

S=1-2+3-4+...+99-100

S=(1-2)+(3-4)+...+(99-100)

S=(-1)+(-1)+...+(-1)

=>S=(-1).50

S=-50

S=-50 nha bạn . 

57/62

Ta có:

$S=\dfrac{5}{1\cdot2\cdot3}+\dfrac{8}{2\cdot3\cdot4}+\dfrac{11}{3\cdot4\cdot5}+\cdots+\dfrac{6026}{2008\cdot2009\cdot2010}$

Nhận thấy tử số có dạng $3n+2$, nên:

$S=\sum_{n=1}^{2008}\dfrac{3n+2}{n(n+1)(n+2)}$

Ta có: $\dfrac{3n+2}{n(n+1)(n+2)}=\dfrac{1}{n(n+1)}+\dfrac{2}{(n+1)(n+2)}$

Do đó:

$S=\left(\dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\cdots+\dfrac1{2008\cdot2009}\right)$

$+2\left(\dfrac1{2\cdot3}+\dfrac1{3\cdot4}+\cdots+\dfrac1{2009\cdot2010}\right)$

Mà: $\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}$

Nên: $S=\left(1-\dfrac1{2009}\right)+2\left(\dfrac12-\dfrac1{2010}\right)$

$=1-\dfrac1{2009}+1-\dfrac1{1005}$

$=2-\dfrac1{2009}-\dfrac1{1005}$

Vì: $\dfrac1{2009}+\dfrac1{1005}>0$

Nên $S<2$

\(S=6^2\left(1^2+2^2+3^2+...+10^2\right)=36.385=13860\)

13860 mình cùng lớp bạn đấy đó bạn biết là ai:))_