1.已知数列an的前n项和为Sn,且Sn=2^n,求通项an;2.已知数列an的前n项和为Sn,且Sn=n^2+3n,求通项an;

来源:学生作业帮助网 编辑:作业帮 时间:2024/03/28 20:52:11
1.已知数列an的前n项和为Sn,且Sn=2^n,求通项an;2.已知数列an的前n项和为Sn,且Sn=n^2+3n,求通项an;

1.已知数列an的前n项和为Sn,且Sn=2^n,求通项an;2.已知数列an的前n项和为Sn,且Sn=n^2+3n,求通项an;
1.已知数列an的前n项和为Sn,且Sn=2^n,求通项an;
2.已知数列an的前n项和为Sn,且Sn=n^2+3n,求通项an;

1.已知数列an的前n项和为Sn,且Sn=2^n,求通项an;2.已知数列an的前n项和为Sn,且Sn=n^2+3n,求通项an;
1、
n>=2
an=Sn-S(n-1)
=2^n-2^(n-1)
=2*2^(n-1)-2^(n-1)
=2^(n-1)
a1=S1=2^1=2
不符合n>=2时的an=2^(n-1)
所以
n=1,an=2
n>=2,an=2^(n-1)
2、
n>=2
an=Sn-S(n-1)
=n²+3n-(n-1)²-3(n-1)
=2n+2
a1=S1=1+3=4
符合an=2n+2
所以an=2n+2

1、
Sn=2^n
Sn-S(n-1)=an=2^n-2^(n-1)=2^(n-1)
2、
an=Sn-S(n-1)=n^2+3n-(n-1)^2-3(n-1)
=(n-n+1)(n+n+1)+3
=2n+4

1.n=1,an=2^1=2,
n≥2,an=Sn-S(n-1)=2^(n-1)
2.n=1,an=1^2+3*1=4,
n≥2,an=Sn-S(n-1)=2n+2
综上,an=2n+2