等差数列an的前n项和为sn,a1=1+根号2,s3=9+3根号2
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等差数列an的前n项和为sn,a1=1+根号2,s3=9+3根号2
(1)设bn=an-根号2(n属于自然数)bn中的部分项bk1,bk2,bkn,恰好组成等比数列.且k1=1,k4=63,求数列kn的通项公式
(1)设bn=an-根号2(n属于自然数)bn中的部分项bk1,bk2,bkn,恰好组成等比数列.且k1=1,k4=63,求数列kn的通项公式
s3=3*(a1+a3)/2=9+3√2
so:a3=5+√2g;so,公差=2
an=2n-1+√2
so:bn=2n-1
bk1=b1=1;bk4=b63=125
so,q=(125/1)开(4-1)次方=5
bkn=2kn-1,bk(n+1)=2k(n+1)-1
q=bk(n+1)/bkn=[ 2k(n+1)-1]/ (2kn-1)=5
so,[k(n+1)-1/2]=5[kn-1/2]
设cn=kn-1/2
c1=k1-1/2=1/2
so,cn=[5^(n-1)]/2
so,kn=[5^(n-1)]/2+1/2
so:a3=5+√2g;so,公差=2
an=2n-1+√2
so:bn=2n-1
bk1=b1=1;bk4=b63=125
so,q=(125/1)开(4-1)次方=5
bkn=2kn-1,bk(n+1)=2k(n+1)-1
q=bk(n+1)/bkn=[ 2k(n+1)-1]/ (2kn-1)=5
so,[k(n+1)-1/2]=5[kn-1/2]
设cn=kn-1/2
c1=k1-1/2=1/2
so,cn=[5^(n-1)]/2
so,kn=[5^(n-1)]/2+1/2
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