求17题过程
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17、(1)an+1=f(an)=an²+an-1/4
an+₁ +1/2=(an+1/2)²
an +1/2=(an-1+1/2)²=(a₁+1/2)²^ⁿˉ¹
a₁+1/2=(3/2)²+3/2-1/4+1/2=4
bn=log₂⁴^²^ⁿˉ¹=2^²^ⁿˉ¹
{bn}为等比数列
(2)cn=bn/(bn-1)(bn+1 -1)
=bn/(bn-1)(4bn -1)
=[1/(bn -1) -1/(4bn -1)]/3
cn=[1/(bn -1) -1/(bn+1 -1)]/3
c₁+c₂+c₃+...+cn=[1/(b₁-1)-1/(bn+1 -1)]/3
an+₁ +1/2=(an+1/2)²
an +1/2=(an-1+1/2)²=(a₁+1/2)²^ⁿˉ¹
a₁+1/2=(3/2)²+3/2-1/4+1/2=4
bn=log₂⁴^²^ⁿˉ¹=2^²^ⁿˉ¹
{bn}为等比数列
(2)cn=bn/(bn-1)(bn+1 -1)
=bn/(bn-1)(4bn -1)
=[1/(bn -1) -1/(4bn -1)]/3
cn=[1/(bn -1) -1/(bn+1 -1)]/3
c₁+c₂+c₃+...+cn=[1/(b₁-1)-1/(bn+1 -1)]/3
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