Question

设数列An的前n项和Sn=2an-2的n次方,求A1,A4

Answer 1

S(n)=2a(n)-2^n

当n=1时,

S(1)=a(1)=2a(1)-2^1

得 a(1)=2；

n=2时,

S(2)=a(1)+a(2)

=2a(2)-2²

解得

a(2)=a(1)+2²=2+4=6；

n=3时,

S(3)=a(1)+a(2)+a(3)

=2a(3)-2³

解得

a(3)=a(1)+a(2)+2³=2+6+8=16；

n=4时,

S(4)=a(1)+a(2)+a(3)+a(4)

=2a(4)-2^4

解得

a(4)=a(1)+a(2)+a(3)+2^4=2+6+16+16=40.