数列{an}的前n项的和Sn=2an-1(n∈N*),数列{bn}满足:b1=3,bn+1=an+bn(n∈N*). (1)求证:数列{an}为等比数列; (2)求数列{bn}的前n项的和Tn.
数列{an}的前n项的和Sn=2an-1(n∈N*),数列{bn}满足:b1=3,bn+1=an+bn(n∈N*).
(1)求证:数列{an}为等比数列;
(2)求数列{bn}的前n项的和Tn.
数学人气:657 ℃时间:2020-01-22 22:36:09
优质解答
(1)∵a
n+1=S
n+1-S
n=(2a
n+1-1)-(2a
n-1),
∴a
n+1=2a
n,
又a
1=S
1=2a
1-1,∴a
1=1≠0,
因此数列{a
n}为公比是2、首项是1的等比数列;
(2)易得
bn+1−bn=2n−1,∴
bn−bn−1=2n−2,
bn−1−bn−2=2n−3,…,
b2−b1=20=1,
以上各式相加得,
bn+1−b1=1+2+3+…+2n−1=2
n-1,
∴
bn+1=2n+2,∴
bn=2n−1+2,
∴T
n=b
1+b
2+…+b
n=2n+
=2
n+2n-1(n∈N*).
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