已知数列{an}的前n项和为Sn=2n^2+n+3,求数列{an}的第10项

来源：百度知道 编辑：UC知道 时间：2024/05/24 12:04:46

∵a10=S10-S9
Sn=2n^2+n+3
∴a10=(2*10^2+10+3)-(2*9^2+9+3)
=39

或者:
∵当n>1时,an=Sn-S(n-1)
=(2*n^2+n+3)-[2*(n-1)^2+(n-1)+3]
=4n-1
∴a10=4*10-1
=39

Sn=2n^2+n+3,
Sn-1=2（n-1)^2+(n-1)+3=2n^2-4n+2+n-1+3
=2n^2+n+3-4n+1
Sn-Sn-1=an=4n-1
所以得：a10=39

∵a10=S10-S9
Sn=2n^2+n+3
∴a10=(2*10^2+10+3)-(2*9^2+9+3)
=39

或者:
∵当n>1时,an=Sn-S(n-1)
=(2*n^2+n+3)-[2*(n-1)^2+(n-1)+3]
=4n-1
∴a10=4*10-1
=39

解：Sn=2n^2+n+3,
Sn-1=2（n-1)^2+(n-1)+3=2n^2-4n+2+n-1+3
=2n^2+n+3-4n+1
Sn-Sn-1=an=4n-1
所以得：a10=39

∵Sn=2n^2+n+3
∴S(n-1)=2(n-1)^2+n-1+3
=2n^2-3n+4
∴an=Sn-S(n-1)
=4n-1
∴a10=4*10-1
=39

A10=S10-S9=(2*10^2+10+3)-(2*9^2+9+3)=39