数列a1=1,an=an+1(1+2an)求证数列an分之一等差数列,若a1a2+a2a3+..+anan+1大于33分
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数列a1=1,an=an+1(1+2an)求证数列an分之一等差数列,若a1a2+a2a3+..+anan+1大于33分之16,求N的取值范
an=an+1(1+2an)
an/(1+2an)=an+1
1/an+1=1/an+2
1/an=1+(n-1)2=2n-1
an=1/(2n-1)
2anan+1=2/(2n-1)*1/(2n+1)=1/(2n-1)-1/(2n+1)
2a1a2=1-1/3
2(a1a2+a2a3+..+anan+1)=1-1/(2n+1)=2n/(2n+1)
a1a2+a2a3+..+anan+1=n/(2n+1)>16/33
n>16
an/(1+2an)=an+1
1/an+1=1/an+2
1/an=1+(n-1)2=2n-1
an=1/(2n-1)
2anan+1=2/(2n-1)*1/(2n+1)=1/(2n-1)-1/(2n+1)
2a1a2=1-1/3
2(a1a2+a2a3+..+anan+1)=1-1/(2n+1)=2n/(2n+1)
a1a2+a2a3+..+anan+1=n/(2n+1)>16/33
n>16
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