Question

已知等差数列满足a2=0 a6+a8=-10 求数列（an）的通项公式 求数列（2的n-1次方分之an）的前n项和

Answer 1

an=a1+(n-1)d,
a1+d=0,a1+5d+a1+7d=-10,
a1=1,d=-1,
Sn-(1/2)Sn=a1+d/2+d/4+d/8+....+d/2^(n-1)-d/2^n
Sn/2=1-[1/2+1/4+...+1/2^(n-1)]+1/2^n =(1/2)^n+(1/2)(n-1)
Sn=(1/2)^(n-1)+(1/2)^(n-2)

Answer 2

a6=a2+4d, a8=a2+6d, a6+a8=2a2+10d=10d=-10, d=-1, a2=a1+d
0=a1-1, a1=1,,,,,an=a1+(n-1)d=1+(n-1)(-1)=1-(n-1)=1-n+1=2-n
Sn=a1/2^0+a2/2^1+---+an/2^n-1=(n-1)/2^(n-2)