平方和等比数列我算出来一个平方等比数列,但不知道怎么求和数列如下：n^2/2^(n-1)〖即是N的平方除以2的N减1次方

s(n)=1^2/2^(1-1)+2^2/2^(2-1)+...+n^2/2^(n-1),
s(n)/2=1^2/2^(2-1)+2^2/2^(3-1)+...+(n-1)^2/2^(n-1)+n^2/2^n,
s(n)/2=s(n)-s(n)/2=1^2/2^(1-1)+[2^2-1^2]/2^(2-1)+...+[n^2-(n-1)^2]/2^(n-1)-n^2/2^n
=1+(2*1+1)/2+(2*2+1)/2^2+...+[2*(n-1)+1]/2^(n-1)-n^2/2^n
=2[1/2+2/2^2+...+(n-1)/2^(n-1)]+[1+1/2+...+1/2^(n-1)]-n^2/2^n
=2t(n)+[(1-1/2^n)/(1-1/2)]-n^2/2^n
=2t(n)+2[1-1/2^n]-n^2/2^n
t(n)=1/2+2/2^2+...+(n-1)/2^(n-1),
t(n)/2=1/2^2+2/2^3+...+(n-2)/2^(n-1)+(n-1)/2^n,
t(n)/2=t(n)-t(n)/2=1/2+(2-1)/2^2+...+[(n-1)-(n-2)]/2^(n-1)-(n-1)/2^n
=1/2+1/2^2+...+1/2^(n-1)-(n-1)/2^n
=[1-1/2^n]/[1-1/2]-1-(n-1)/2^n
=2[1-1/2^n]-1-(n-1)/2^n,
s(n)=2*s(n)/2=2*[2t(n)+2[1-1/2^n]-n^2/2^n]
=4t(n)+4[1-1/2^n]-n^2/2^(n-1)
=8*t(n)/2+4[1-1/2^n]-n^2/2^(n-1)
=8{2[1-1/2^n]-1-(n-1)/2^n}+4[1-1/2^n]-n^2/2^(n-1)
=16[1-1/2^n]-8-8(n-1)/2^n+4[1-1/2^n]-n^2/2^(n-1)
=12-[10+4(n-1)+n^2]/2^(n-1)