Question

若数列{an}的前n项和sn=n^+3n+1求a1+a3+a5+······+a21的值,谢谢！

Answer 1

分析 由Sn=n2+3n+1可得

Sn-Sn-1=n2+3n+1-(n-1)2-3(n-1)-1

=2(n+1)

=4+(n-1)·2 (n≥2)

∵a1=S1=5，∴a2=S2-a1=6

所以数列{an}是除a1以外是等差数列的数列，其首项为6，公差为2。则

a1+a3+a5+…+a21

=a1+(a3+a5+…+a21)

=5+[10*(a3+a21)/2]

=5+[10*(8+44)/2]

=265

Answer 2

a1=s1=1^2+3*1+1=5

sn=n^2+3n+1
s(n-1)=(n-1)^2+3(n-1)+1
=n^2-2n+1+3n-3+1
=n^2+n-1

an=sn-s(n-1)
=n^2+3n+1-(n^2+n-1)
=n^2+3n+1-n^2-n+1
=2n+2

an=2n+2(n>=2)

a3=2*3+2=8
a21=2*21+2=44

a1+a3+a5+······+a21
=a1+(a3+a21)*10/2
=5+(8+44)*5
=5+52*5
=5+260
=265