数列{an}中,an=1/n(n+1),{an}的前n项和为2009/2010,则项数n为A2008 B2009 C2010 C2011

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数列{an}中,an=1/n(n+1),{an}的前n项和为2009/2010,则项数n为A2008B2009C2010C2011数列{an}中,an=1/n(n+1),{an}的前n项和为2009/

数列{an}中,an=1/n(n+1),{an}的前n项和为2009/2010,则项数n为A2008 B2009 C2010 C2011
数列{an}中,an=1/n(n+1),{an}的前n项和为2009/2010,则项数n为
A2008 B2009 C2010 C2011

数列{an}中,an=1/n(n+1),{an}的前n项和为2009/2010,则项数n为A2008 B2009 C2010 C2011
an=[(n+1)-n]/n(n+1)
=(n+1)/n(n+1)-1/n(n+1)
=1/n-1/(n+1)
所以Sn=1-1/2+1/2-1/3+……+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
=2009/2010
所以n=2009

sn=1-1/(n+1)
=n/(n+1)=2009/2010
所以n=2009
选B

B2009
sn=1/1*2+1/2*3+...+1/n(n+1)
=1-1/2+1/2-1/3+...+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
n/(n+1)=2009/2010
n=2009

Sn=1-1/(n+1)=2009/2010
所以1/(n+1)=1/2010
所以n=2009

an=1/n(n+1)可以变形为(1/n)-(1/n+1)这样前n项和就变为了1-(1/n+1)即n/n+1这样可得n为2009
选择B