数列{an}满足a1=1,Sn+1=Sn+(n+1)(n∈N) (1)用an表示an+1 (2)证明:数列{(an)+1}是等比数列 (3)求an和Sn
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![数列{an}满足a1=1,Sn+1=Sn+(n+1)(n∈N) (1)用an表示an+1 (2)证明:数列{(an)+1}是等比数列 (3)求an和Sn](/uploads/image/z/3926880-0-0.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2CSn%2B1%3DSn%2B%28n%2B1%29%28n%E2%88%88N%29+%281%29%E7%94%A8an%E8%A1%A8%E7%A4%BAan%2B1+%282%29%E8%AF%81%E6%98%8E%EF%BC%9A%E6%95%B0%E5%88%97%7B%28an%29%2B1%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97+%EF%BC%883%EF%BC%89%E6%B1%82an%E5%92%8CSn)
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数列{an}满足a1=1,Sn+1=Sn+(n+1)(n∈N) (1)用an表示an+1 (2)证明:数列{(an)+1}是等比数列 (3)求an和Sn
数列{an}满足a1=1,Sn+1=Sn+(n+1)(n∈N) (1)用an表示an+1 (2)证明:数列{(an)+1}是等比数列 (3)求an和Sn
数列{an}满足a1=1,Sn+1=Sn+(n+1)(n∈N) (1)用an表示an+1 (2)证明:数列{(an)+1}是等比数列 (3)求an和Sn
1、S(n+1)=2Sn+(n+1),所以,当n≥2时,有:Sn=2S(n-1)+n,两式相减,得:a(n+1)=2an+1;
2、a(n+1)+1=2an+2=2(an+1),即[a(n+1)+1]/[an+1]=2=常数,所以数列{(an)+1}是等比数列.
3、数列{(an)+1}是等比数列,则an+1=[(a1)+1]×2^(n-1),即an=2^(n-1)-1.再利用分组求和的方法求出Sn=2^n-1-n.
(1)a(n+1)=S(n+1)-Sn=n+1;
an=Sn-S(n-1)=n;
所以a(n+1)=(an)+1.
(2)等差数列吧?
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