Question

已知数列an的前n项和sn=(n^2+n)/2 设bn=2^an+(-1)^n×an 求 数列

n的前n项和

Answer 1

Sn = (n^2+n)/2
n=1, a1=1
for n>=2
an = Sn -S(n-1)
=(1/2)( 2n-1 +1)
=n
bn =2^(an) + (-1)^n .an
=2^n + (-1)^n .n
Tn = b1+b2+...+bn
if n is odd
Tn = 2(2^n -1) - n + (n-1)/2
=2(2^n -1) - (n+1)/2
if n is even
Tn =2(2^n -1) + n/2

追问

分类之后的计算过程可以详细点吗

分类之后的计算过程可以详细点吗

追答

bn= 2^n +(-1)^n.n
let
dn =2^n
d1+d2+...+dn = 2(2^n-1)

en =(-1)^n.n
e1+e2+...+en= -1+2+...+(-1)^n
if n is odd
e1+e2+...+en
= -1+2+...+(n-1) -n
= [-1+2]+[-3+4] ...+(-(n-2)+(n-1)] -n
= (n-1)/2 -n
=-(n+1)/2

if n is even
e1+e2+...+en
= -1+2+...-(n-1) +n
= [-1+2]+[-3+4] ...+[-(n-1)+n]
= n/2

bn= dn +en
ie

Tn = b1+b2+...+bn
if n is odd
Tn = 2(2^n -1) - n + (n-1)/2
=2(2^n -1) - (n+1)/2
if n is even
Tn =2(2^n -1) + n/2