Question

数列求和：Sn=（2n-1）除以2的n-1次方

Answer 1

sn=(2n-1)/2^(n-1)

s(n-1)=(2n-3)/2^(n-2)

sn-s(n-1)=(2n-1)/2^(n-1)-(2n-3)/2^(n-2)

an=(2n-1)/2^(n-1)-(2n-3)/2^(n-2)

an=(2n-1)/2^(n-1)-2*(2n-3)/2^(n-1)

an=[(2n-1)-2*(2n-3)]/2^(n-1)

an=[2n-1-4n+6]/2^(n-1)

an=(5-2n)/2^(n-1)