若数列{an}的通项公式是an=3-n+(-2)-n+1,则
|
a1+a2+…+an=(3-1+1)+[3-2+(-2)-1]+[3-3+(-2)-2]+…+[3-n+(-2)-n+1
=(3-1+3-2+…+3-n)+…+[1+(-2)-1+(-2)-2+…+(-2)-n+1]
=
+3-1(1-3-n) 1-3-1
=1•[1-(-
)n]1 2 1-(-2)-1
+1- 1 3n 2
,1-(-
)n1 2 3 2
所以
(a1+a2+…+an)=lim n→∞
[lim n→∞
+1- 1 3n 2
]=1-(-
)n1 2 3 2
.7 6
故答案为:
.7 6