记公差d≠0的等差数列{an}的前n项和为Sn,一直a1=2+根号2,S3=12+3根号2.记bn=an-根号2,若自然
记公差d≠0的等差数列{an}的前n项和为Sn,一直a1=2+根号2,S3=12+3根号2.记bn=an-根号2,若自然数n1,n2…
记bn=an-根号2,若自然数n1,n2…,nk…,满足1≤n1≤n2≤……≤nk≤……,并且bn1,bn2,…,bnk,…,成等比数列,其中n1=1,n2=3,求nk（用k表示）
记bn=an-根号2,若自然数n1,n2…,nk…,满足1≤n1≤n2≤……≤nk≤……,并且bn1,bn2,…,bnk,…,成等比数列,其中n1=1,n2=3,求nk（用k表示）
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a(n)=a+(n-1)d=2+2^(1/2) + (n-1)d.
s(n) = na + n(n-1)d/2 = n[2+2^(1/2)] + n(n-1)d/2,
12 + 3(2)^(1/2) = s(3) = 3[2+2^(1/2)] + 3d,d = 2.
a(n) = 2+2^(1/2) + 2(n-1).
b(n) = a(n) - 2^(1/2) = 2+2(n-1)=2n.
b[n(k)] = 2n(k) = b[n(1)]*q^(k-1) = 2n(1)*q^(k-1) = 2q^(k-1),
b[n(2)] = 2n(2) = 6 = 2q,q = 3.
b[n(k)] = 2n(k) = 2q^(k-1) = 2*3^(k-1),
n(k) = 3^(k-1)
s(n) = na + n(n-1)d/2 = n[2+2^(1/2)] + n(n-1)d/2,
12 + 3(2)^(1/2) = s(3) = 3[2+2^(1/2)] + 3d,d = 2.
a(n) = 2+2^(1/2) + 2(n-1).
b(n) = a(n) - 2^(1/2) = 2+2(n-1)=2n.
b[n(k)] = 2n(k) = b[n(1)]*q^(k-1) = 2n(1)*q^(k-1) = 2q^(k-1),
b[n(2)] = 2n(2) = 6 = 2q,q = 3.
b[n(k)] = 2n(k) = 2q^(k-1) = 2*3^(k-1),
n(k) = 3^(k-1)
1年前
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