已知An=n^2乘2^n 求Sn 在线等！！！

An=n^2 *2^n
A(n-1) =(n-1)^2 *2^(n-1)
An-2A(n-1)=2^n *(n^2-(n-1)^2) =2^n *(2n-1)
Sn=A1+A2+A3+...+A(n-1)+A(n)
Sn-2Sn=A1+(A2-2A1)+(A3-2A2)+....+(An-2A(n-1)-2A(n)
=A1+3*2^2+5*2^3+......+(2n-1)*2^n -2A(n)
令t=3*2^2+5*2^3+.....+(2n-1)*2^n
2t=3*2^3+5*2^4+.....+(2n-3)*2^n +(2n-1)*2^(n+1)
2t-t=t=-3*2^2+(3-5)2^3+(5-7)2^4+.....+(2n-3-2n+1)2^n+(2n-1)*2^(n+1)
=-2(2^3+2^4+.....+2^n) +(2n-1)2^(n+1)-3*2^2
=-2(2+2^2+2^3+...+2^n) +4+8+(2n-1)2^(n+1)-12
=-2 *2(1-2^n)/(1-2) +(2n-1)2^(n+1)
=4-4*2^n +(2n-1)2^(n+1)
所以
-Sn=Sn-2Sn=A1+t-2An
=2+4-4*2^n+(2n-1)2^(n+1)-2*n^2 *2^n
=6+(2n-3)2^(n+1)-n^2*2^(n+1)
Sn=-6-(2n-3)2^(n+1)+n^2*2^(n+1)
=(n^2 -2n+3)*2^(n+1) -6

an =n^2 . 2^n
= 2 ( n^2 .2^(n-1) )
consider
1+x+x^2+...+x^n = (x^(n+1) -1)/(x-1)
1+2x+3x^2+...+nx^(n-1)= [(x^(n+1) -1)/(x-1)]'
= [nx^(n+1) -(n+1)x^n + 1]/(x-1)^2
x+2x^2+3x^3+...+nx^n = x[nx^(n+1) -(n+1)x^n + 1]/(x-1)^2
1+2^2x+3^2x^2+...+n^2x^(n-1)
= [x[nx^(n+1) -(n+1)x^n + 1]/(x-1)^2]'
= { (x-1)^2[ n(n+2)x^(n+1)- (n+1)^2.x^n + 1] - 2x(x-1)[nx^(n+1) -(n+1)x^n + 1] } /(x-1)^4
put x= 2
∑(i:1->n)[ i^2 .( 2^(i-1)]
= [ n(n+2). 2^(n+1)- (n+1)^2 .2^n + 1] - 4[n.2^(n+1) -(n+1)2^n + 1]
= -3 + 2^n . { 2n(n+2) - (n+1)^2 - 8n + 4(n+1) }
= -3 + (n^2-2n+3). 2^n
Sn = a1+a2+..+an
= 2∑(i:1->n)[ i^2 .( 2^(i-1)]
= 2[-3 + (n^2-2n+3). 2^n]

你知道错位相减还问- -/// 是n^2*2^n这样？ 不是应该是N*（2^n）才是么 等差*等比·