已知数列｛an｝（n∈N*），若a1=1，an+1+an=（1/2）^n，则lima2n=______,用 累加法 如何计算？
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a(n+1)+an=(1/2)^n
a(n+1) - (2/3).(1/2)^(n+1) = - [ an -(2/3) (1/2)^n ]
=>{ an -(2/3) (1/2)^n } 是等比数列, q=-1
an -(2/3) (1/2)^n =(-1)^(n-1) . [a1 -(2/3) (1/2)]
= (2/3).(-1)^(n-1)
an =(2/3) (1/2)^n + (2/3).(-1)^(n-1)
a(2n)= (2/3) (1/2)^(2n) - 2/3
lim(n->∞) a(2n)
=lim(n->∞) [ (2/3) (1/2)^(2n) - 2/3 ]
=-2/3
a(n+1) - (2/3).(1/2)^(n+1) = - [ an -(2/3) (1/2)^n ]
=>{ an -(2/3) (1/2)^n } 是等比数列, q=-1
an -(2/3) (1/2)^n =(-1)^(n-1) . [a1 -(2/3) (1/2)]
= (2/3).(-1)^(n-1)
an =(2/3) (1/2)^n + (2/3).(-1)^(n-1)
a(2n)= (2/3) (1/2)^(2n) - 2/3
lim(n->∞) a(2n)
=lim(n->∞) [ (2/3) (1/2)^(2n) - 2/3 ]
=-2/3
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