Question

高三数学数列

已知数学(an)满足a1=2,an+1=2an-n+1问题①，当数列(bn)满足bn=an-n时，试判断{bn}是等差还是等比数列，并证明，②求an的前n项和Sn

Answer 1

a(n+1)=2an-n+1
a(n+1)=2(an-n)+1+n
a(n+1)-(n+1)=2(an-n)
[a(n+1)-(n+1)]/(an-n)=2
an-n是以2为公比的等比数列
an-n=(a1-1)q^(n-1)
an-n=(2-1)*2^(n-1)
an-n=2^(n-1)
an=2^(n-1)+n

sn=a1+a2+....+an
=2^0+1+2^1+2+........+2^(n-1)+n
=2^0+2^1+........+2^(n-1)+1+2+........+n
=1*(1-2^n)/(1-2)+n(n+1)/2
=2^n-1+n(n+1)/2
=2^n+(n^2+n-2)/2

Answer 2

a(n+1)=2(an-n)+1+n
a(n+1)=2an-n+1
a(n+1)-(n+1)=2(an-n)
[a(n+1)-(n+1)]/(an-n)=2
an-n是以2为公比的等比数列
an-n=(a1-1)q^(n-1)
an-n=(2-1)*2^(n-1)
an-n=2^(n-1)
an=2^(n-1)+n