若数列bn的通项公式为bn=n^2*2^n,则其前n项和Tn为?用错位相减法求、答案请尽可能详细、谢谢.

Tn=b1+b2+b3+...+bn=1²×2^1+2²×2^2+3²×2^3+...+n²×2^n
2Tn=1²×2^2+2²×2^3+...+(n-1)²×2^n+n²×2^(n+1)
Tn-2Tn=-Tn=1²×2+(2²-1²)×2^2+(3²-2²)×2^3+...+[n²-(n-1)²]×2^n-n²×2^(n+1)
=2+(2+1)(2-1)×2^2+(3+2)(3-2)×2^3+...+[(n+n-1)(n-n+1)]×2^n-n²×2^(n+1)
=1×2^1+3×2^2+5×2^3+...+(2n-1)×2^n -n²×2^(n+1)
令Cn=1×2^1+3×2^2+5×2^3+...+(2n-1)×2^n
则2Cn=1×2^2+3×2^3+...+(2n-3)×2^n+(2n-1)×2^(n+1)
Cn-2Cn=-Cn=2^1+2×2^2+2×2^3+...+2×2^n-(2n-1)×2^(n+1)
=2+2(2^2+2^3+...+2^n)-(2n-1)×2^(n+1)
=2+2×4×[2^(n-1)-1]/(2-1)-(2n-1)×2^(n+1)
=2+8×2^(n-1)-8-(2n-1)×2^(n+1)
=(3-2n)×2^(n+1)-6
Cn=(2n-3)×2^(n+1) +6
-Tn=1×2^1+3×2^2+5×2^3+...+(2n-1)×2^n -n²×2^(n+1)
=(2n-3)×2^(n+1)+6-n²×2^(n+1)
=(2n-3-n²)×2^(n+1) +6
Tn=(n²-2n+3)×2^(n+1) -6
提示：两次使用错位相减法.