Question

已知数列an=（2n+1）*2^n次方，求前n项和Sn

Answer 1

解：
Sn=3·2+5·2²+7·2³+...+(2n+1)·2ⁿ
2Sn=3·2²+5·2³+...+(2n-1)·2ⁿ+(2n+1)·2ⁿ⁺¹
Sn-2Sn=-Sn=3·2+2·2²+...+2·2ⁿ-(2n+1)·2ⁿ⁺¹
=2+2²+...+2ⁿ⁺¹-(2n+1)·2ⁿ⁺¹
=2·(2ⁿ⁺¹-1)/(2-1) -(2n+1)·2ⁿ⁺¹
=(1-2n)·2ⁿ⁺¹-2
Sn=(2n-1)·2ⁿ⁺¹+2

Answer 2

解：因为an=2n+1
所以{an}是等差数列
所以sn=n(a1+an)/2=n(3+2n+1)/2=n(n+2)
所以1/sn=1/n(n+2)=[1/n-1/(n+2)]/2
所以数列{1/sn}的前n项和tn=s1+s2+...+sn
=[1/1-1/(1+2)]/2+[1/2-1/(2+2)]/2+...+[1/n-1/(n+2)]/2
=[1+1/2-1/(n+1)-1/(n+2)]/2
=3/4-(2n+3)/[2(n+1)(n+2)]