设数列{an}的前n项和Sn=n^2+2n (1)求证{an}是等差数列,并求它的通项公式 (2)设bn=2^an,求证{bn}是等比数列,并求它的前n项和

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设数列{an}的前n项和Sn=n^2+2n(1)求证{an}是等差数列,并求它的通项公式(2)设bn=2^an,求证{bn}是等比数列,并求它的前n项和设数列{an}的前n项和Sn=n^2+2n(1)

设数列{an}的前n项和Sn=n^2+2n (1)求证{an}是等差数列,并求它的通项公式 (2)设bn=2^an,求证{bn}是等比数列,并求它的前n项和
设数列{an}的前n项和Sn=n^2+2n (1)求证{an}是等差数列,并求它的通项公式 (2)设bn=2^an,
求证{bn}是等比数列,并求它的前n项和

设数列{an}的前n项和Sn=n^2+2n (1)求证{an}是等差数列,并求它的通项公式 (2)设bn=2^an,求证{bn}是等比数列,并求它的前n项和
证明:(1)an=Sn-Sn-1=n^2+2n -(n-1)^2-2(n-1)=2n+1(n≥1)
又S1=a1=3,an-an-1=2
所以{an}是是以公差为2的等差数列,通项公式为an=2n+1(n≥1).
(2)bn=2^(2n+1),所以bn-1=2^(2n-1)
所以bn/bn-1=4
所以b1=8,bn=8×4^(n-1) (n≥1)
所以{an}是是以公比为4的等比数列,通项公式为bn=8×4^(n-1) (n≥1)
Sn=8×[(4^n)-1]/3.

a1=s1=1^2+2*1=3
sn=n^2+2n
s(n-1)=(n-1)^2+2(n-1)
=n^2-2n+1+2n-2
=n^2-1
an=sn-s(n-1)
=n^2+2n-(n^2-1)
=n^2+2n-n^2+1
=2n+1
所以{an}是等差数列,an=2n+1
bn=2^an
=2^(2n+1)...

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a1=s1=1^2+2*1=3
sn=n^2+2n
s(n-1)=(n-1)^2+2(n-1)
=n^2-2n+1+2n-2
=n^2-1
an=sn-s(n-1)
=n^2+2n-(n^2-1)
=n^2+2n-n^2+1
=2n+1
所以{an}是等差数列,an=2n+1
bn=2^an
=2^(2n+1)
b(n-1)=2^a(n-1)
=2^(2n-1)
bn/b(n-1)
=2^(2n+1)/2^(2n-1)
=2^[(2n+1)-(2n-1)]
=2^2
=4
所以{bn}是以4为公比的等比数列
sn=b1+b2+...+bn
=2^3+2^5+...+2^(2n+1)
=8*(1-4^n)/(1-4)
=8*(1-2^2n)/(-3)
=2^(2n+3)/3-8/3

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