数列{an}的前n项和为Sn,已知a1=1/2,Sn=n^2-n(n-1)(n=1,2,3.)写出Sn与Sn+1的递推关系式(n大于等于2并求Sn关于n的表达式设fn=(Sn/n)X^n+1,bn=f'n(p)(p属于R),求数列{bn}的前n项和Tn
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![数列{an}的前n项和为Sn,已知a1=1/2,Sn=n^2-n(n-1)(n=1,2,3.)写出Sn与Sn+1的递推关系式(n大于等于2并求Sn关于n的表达式设fn=(Sn/n)X^n+1,bn=f'n(p)(p属于R),求数列{bn}的前n项和Tn](/uploads/image/z/9317947-67-7.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A5a1%3D1%2F2%2CSn%3Dn%5E2-n%28n-1%29%28n%3D1%2C2%2C3.%29%E5%86%99%E5%87%BASn%E4%B8%8ESn%2B1%E7%9A%84%E9%80%92%E6%8E%A8%E5%85%B3%E7%B3%BB%E5%BC%8F%EF%BC%88n%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8E2%E5%B9%B6%E6%B1%82Sn%E5%85%B3%E4%BA%8En%E7%9A%84%E8%A1%A8%E8%BE%BE%E5%BC%8F%E8%AE%BEfn%3D%28Sn%2Fn%29X%5En%2B1%2Cbn%3Df%27n%28p%29%28p%E5%B1%9E%E4%BA%8ER%EF%BC%89%2C%E6%B1%82%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn)
数列{an}的前n项和为Sn,已知a1=1/2,Sn=n^2-n(n-1)(n=1,2,3.)写出Sn与Sn+1的递推关系式(n大于等于2并求Sn关于n的表达式设fn=(Sn/n)X^n+1,bn=f'n(p)(p属于R),求数列{bn}的前n项和Tn
数列{an}的前n项和为Sn,已知a1=1/2,Sn=n^2-n(n-1)(n=1,2,3.)写出Sn与Sn+1的递推关系式(n大于等于2
并求Sn关于n的表达式
设fn=(Sn/n)X^n+1,bn=f'n(p)(p属于R),求数列{bn}的前n项和Tn
数列{an}的前n项和为Sn,已知a1=1/2,Sn=n^2-n(n-1)(n=1,2,3.)写出Sn与Sn+1的递推关系式(n大于等于2并求Sn关于n的表达式设fn=(Sn/n)X^n+1,bn=f'n(p)(p属于R),求数列{bn}的前n项和Tn
1.Sn=n*an-n(n-1)
Sn-1=(n-1)an-1-(n-2)(n-1) n>1
前式减后式
an=n*an-(n-1)an-1-2(n-1)
(n-1)*an-(n-1)an-1-2(n-1)=0
(n-1)(an-an-1-2)=0 n>1
an-an-1=2 n>1
数列(an)是公差为2的等差数列
an=1/2+2(n-1)=2n-3/2
S1=a1=1/2
S2=1/2+1/2+2=3
S3=1/2+1/2+2+1/2+4=15/2
Sn=(1/2+2n-3/2)n/2
=(2n-1)n/2
2.Fn(x)=(Sn/n)x^(n+1)=nx^(n+1)/(n+1)
F'n(p)=n(n+1)p^n/(n+1)=np^n=Bn
若p=1,则Bn=n,则Tn=n(n+1)/2;
若p≠1,这是个很熟悉的关系式,利用错位相减:
Tn=p+2p²+……+np^n
pTn=p²+2p^3+……+np^(n+1)
两式相减=(p-1)Tn=np^(n+1)-(p+p²+……+p^n)=np^(n+1)-p(1-p^n)/(1-p)
Tn=[np^(n+1)-p(1-p^n)/(1-p)]/(p-1)