an+a(n+1) =1/2^n
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 -1/3 )
an =(2/3) (1/2^n) + (-1)^(n-1) . ( a1 -1/3 )
Sn
=a1+a2+...+an
=(2/3)(1- 1/2^n) + ( a1 -1/3 )∑(i:1->n) (-1)^(i-1)
lim(n->∞) Sn =A
=>
a1- 1/3 =0
a1=1/3
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 -1/3 )
an =(2/3) (1/2^n) + (-1)^(n-1) . ( a1 -1/3 )
Sn
=a1+a2+...+an
=(2/3)(1- 1/2^n) + ( a1 -1/3 )∑(i:1->n) (-1)^(i-1)
lim(n->∞) Sn =A
=>
a1- 1/3 =0
a1=1/3
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