高中等差数列前n项和问题、谢

楼上的明显算错了,代入A1 A2就知道了
An=(3n-1)*2^n
Sn=A1+A2+A3+.......(3n-1)*2^n
S(n+1)=A1+A2+A3+.....(3n-1)*2^n+(3n+2)*2^(n+1)
用错位相减法S(n+1)-Sn
得到: A1+(A2-A1)+(A3-A2)+(A4-A3)......[(3n+2)*2^(n+1)-(3n-1)*2^n]
新数列:A(n+1)-An=(3n+5)*2^n=(3n-1)*2^n+6*2^n=An+6*2^n
所以:S(n+1)-Sn=A1+Sn+S(6*2^n)=A1+Sn+6[2^(n+1)-2]=4+Sn+6*2^(n+1)-12
S(n+1)-Sn=A(n+1)=(3n+2)*2^(n+1)
由上面两式可得Sn=(3n-4)*2^(n+1)+8

设前n项和Sn=4+20+64+........+(3n-1)*2^n
∵Sn=(3*1*2-2)+(3*2*2²-2²)+(3*3*2³-2³)+.......+(3*n*2^n-2^n)
=3(1*2+2*2²+3*2³+......+n*2^n)-(2+2²+2³+......+2^n)
设An=1*2+2*2²+3*2³+......+n*2^n..........(1)
Bn=2+2²+2³+......+2^n...........(2)
∴Sn=3An-Bn
∵Bn=2(1-2^n)/(1-2)=2^(n+1)-2
2An=1*2²+2*2³+3*2^4+......+n*2^(n+1)...........(3)
∴(3)-(1)得An=2+2²+2³.......+2^n-n*2^(n+1)
=2+n*2^(n+1)-2^(n+1)
∴Sn=3An-Bn
=3[2+n*2^(n+1)-2^(n+1)]