Question

高一数学。已知an=2^n+n-2 求Sn 已知an=n*(1/3)^n 求Sn

已知an=2^n+n-2 求Sn

已知an=n*(1/3)^n 求Sn

Answer 1

an=2^n+n-2
Sn=2^1+1-2+2^2+2-2+....+2^n+n-2
=(2^1+2^2+2^3+.....+2^n)+(1+2+3+....+n)-2*n
=2*(1-2^n)/(1-2)+n(n+1)/2-2n
=2*(2^n-1)+(n^2+n-4n)/2
=2^(n+1)-2+(n^2-3n)/2
=2^(n+1)+(n^2-3n)/2-2

an=n*(1/3)^n
Sn=1*(1/3)^1+2*(1/3)^2+3*(1/3)^3+.....+n*(1/3)^n
1/3*Sn=1*(1/3)^2+2*(1/3)^3+3*(1/3)^4+.....+n*(1/3)^(n+1)
Sn-1/3*Sn=(1/3)^1+(1/3)^2+(1/3)^3+.....+(1/3)^n+n*(1/3)^(n+1)
2/3*Sn=[1-(1/3)^n]/2+n*(1/3)^(n+1)
2/3*Sn=1/2-(1/3)^n/2+n*(1/3)^(n+1)
2/3*Sn=1/2-(1/3)^n/2+n/3*(1/3)^n
2/3*Sn=1/2+n/3*(1/3)^n-(1/3)^n/2
2/3*Sn=1/2+(n/3-1/2)*(1/3)^n
Sn=3/4+3(n/6-1/4)*(1/3)^n
Sn=3/4+(n/6-1/4)*(1/3)^(n-1)

Answer 2

1：主要是考察等比数列和等差数列的前n项和公式；
Sn=2*(1-2^n)/(1-2)+(-1+n-2)*n/2=2^(n+1)-n^2/2-3n/2-2
2:主要考察等差数列比上等比数列的前n项求法；此类办法是乘以等比数列的公比然后相减可得；
即设Sn=1/3^1+2/3^2+3/3^3+......+n/3^n
乘以公比1/3Sn= 1/3^2+2/3^3+......+(n-1)/3^n+n/3^(n+1)
两式相减可得：2/3Sn=1/3+1/3^2+....+1/3^n -n/3^(n+1)
2/3Sn=1/3*(1-(1/3)^n)/(1-1/3)-n/3^(n+1)
则Sn=(3^(n+1)-2n-3)/(4*3^n)