已知数列{An}满足2an+1=an+an+2,它的前n项和为Sn,且a3=5
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已知数列{An}满足2a(n+1)=an+a(n+2),它的前n项和为Sn,且a3=5,s6=36,(1)求an;(2)已知等比数列bn满足b1+b2=1+a,b4+b5=a^3+a^4(a不等于-1),设数列{an*bn}的前n项和为Tn,求Tn
2a(n+1)=an+a(n+2)
=>
a(n+2)-a(n+1)=a(n+1)-an
这是等比数列
S6=a1+a6+a2+a5+a3+a4=3a3+3a4=36
=>
a4=7
d=a4-a3=2
a1=1
=>
an = 1+2(n-1)
=2n-1
(2)
(b4+b5)=q^3(b1+b2)
=>
a^3(a+1)=q(a+1)
a+1≠0
=>
q=a
=>
b1+b2=b1(1+a)=1+a
=>
b1=1
bn=a^(n-1)
anbn=(2n-1)a^(n-1)
求Tn用错位相减法
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