等差数列求通项公式
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an =a1+(n-1)d
a2=-4/5
a1+d =-4/5 (1)
1/(1+an) -1/(1+a(n-1)) =2
=>
{1/(1+an) } 是等差数列, 公差=2
1/(1+an) - 1/(1+a1) = 2(n-1)
n=2
1/(1+a2) - 1/(1+a1) = 2
1/(1-4/5) -1/(1+a1) = 2
1/(1+a1) = 3
3+3a1=1
a1= -2/3
from (1)
a1+d =-4/5
-2/3 +d =-4/5
d= (10-12)/15 = -2/15
an = a1+(n-1)d = -2/3 -(2/15)(n-1)
a2=-4/5
a1+d =-4/5 (1)
1/(1+an) -1/(1+a(n-1)) =2
=>
{1/(1+an) } 是等差数列, 公差=2
1/(1+an) - 1/(1+a1) = 2(n-1)
n=2
1/(1+a2) - 1/(1+a1) = 2
1/(1-4/5) -1/(1+a1) = 2
1/(1+a1) = 3
3+3a1=1
a1= -2/3
from (1)
a1+d =-4/5
-2/3 +d =-4/5
d= (10-12)/15 = -2/15
an = a1+(n-1)d = -2/3 -(2/15)(n-1)
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