设数列{an}的前n项和为Sn,满足2Sn=an+1-2^(n+1)+1,且a1,a2+5.a3成等差数列,求数列{an
设数列{an}的前n项和为Sn,满足2Sn=an+1-2^(n+1)+1,且a1,a2+5.a3成等差数列,求数列{an}的通项公式；证明：对一切正整数n,有1/a1+1/a2+...1/an
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a1,a2+5,a3成等差数列
a1+a3 = 2(a2+5)
2a2-a3= a1-10 (1)
2Sn=a(n+1)-2^(n+1)+1
n=2
2S2 = a3- 7
2(a1+a2) = a3-7
2a2-a3 = -2a1-7 ...
a1+a3 = 2(a2+5)
2a2-a3= a1-10 (1)
2Sn=a(n+1)-2^(n+1)+1
n=2
2S2 = a3- 7
2(a1+a2) = a3-7
2a2-a3 = -2a1-7 ...
1年前
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