Question

已知数列{an}的通项公式an=2n+1/2^n求，求数列{an}的前n项和Sn 详细点

Answer 1

sn=2*1+1/2^1+2*2+1/2^2+2*3+1/2^3+.....+2^n+1/2^n
=2*(1+2+3+...+n)+1/2^1+1/2^2+1/2^3+...+1/2^n
=2*(1+n)n/2+1/2*[1-(1/2)^n]/(1-1/2)
=n(n+1)+1/2*[1-(1/2)^n]/(1/2)
=n(n+1)+1-(1/2)^n
=n^2+n-(1/2)^n+1

Answer 2

分开算 an=bn+cn
其中：
bn=2n 则Sbn=n(n+1)
cn=1/2^n 则Scn=1/2(1-1/2^n)/(1-1/2)=1-1/2^n
于是Sn=n(n+1)+1-1/2^n

Answer 3

a1=2+1/2^1
a1+a2+.....an
=2+1/2^1+2*2+1/2^2+.....+2n+1/2^n
=2+2*2+.....+2n+ 1/2^1+1/2^2+.....+1/2^n
=(2+2n)*n/2+(1/2^1+1/1/2^n)*n/2
=n(n+5/4)+n/2^(n+1)

Answer 4

分别求2n、1/2^n的前n项和再相加
∑2n=2+4+6+...+2n=n^2+n
∑1/2^n=1/2+1/2^2+1/2^3+...+1/2^n=1-1/2^n
所以：an=n^2+n+1-1/2^n

Answer 5

只就是一个等差加一个等比数列的求和 只是放在一起了