数列{an}是首项为a,公比为t的等比数列,bn=1+a1+a2+......+an,cn=2+b1+b2+....+bn,求bn,cn

数列{an}是首项为a,公比为t的等比数列,
an = at^(n-1), n = 1,2,...

bn=1+a1+a2+......+an

=1 + a + at + ... + at^(n-1)

=1 + a[1 + t + ... + t^(n-1)]

=1 + a(1 - t^n)/(1 - t), n = 1,2,...

cn=2+b1+b2+....+bn

=2 + (1+a) + [1+a(1+t)] +... + [1 + a(1 - t^n)/(1 - t)]

=2 + n + a[1 + (1+t) + ... + (1-t^n)/(1-t)]

=2+n + a[(1-t) + (1-t^2) + ... +(1-t^n)]/(1-t)
=2+n + a[n-(t+t^2+...+t^n)]/(1-t)

=2+n + a{n-t[1+t+...+t^(n-1)]}/(1-t)

=2+n + a[n - t(1-t^n)/(1-t)]/(1-t)

=2+n + a[n(1-t) -t(1-t^n)]/[(1-t)^2],
n = 1,2,...