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