hu .....hu ...làm ơn làm phươc giải giúp bài nay đi hu......hu

L

linh_thuy

Ta có 2S=1/2+2/2^2 +3/2^3 +...+n/2^n+...+2006/2^2006.
S=2S-S=(1/2+2/2^2 +3/2^3 +...+n/2^n+...+2006/2^2006)
- (1/2^2 +2/2^3 +...+n/2^(n+1)+...+2006/2^2007)
=1/2 +1/2^2+1/2^3+...+1/2^(n+1)+...+1/2^2006-2006/2^2007
=(1/2)(1-(1/2)^2006)/(1-1/2)- 2006/2^2007
=1-1/2^2006-2006/2^2007
=1-2008/2^2007
Vậy S<1
 
Last edited by a moderator:
You must log in or register to reply here.
Top Bottom