Question

已知数列{An}的前n项和为Sn,且a1=1,An+1=2Sn.求数列{An}的通项公式An.设Bn=nAn,求{Bn}数列的前n项和Tn

Answer 1

1An+1=2SnAn=2Sn-1An+1-An=2(Sn-Sn-1)=2AnAn+1/An=3=qAn=A1q^(n-1)=3^(n-1)2Bn=nAn=n*3^(n-1)Tn=n*3^(n-1)+(n-1)*3^(n-2)+...+6+13Tn=n*3^n+(n-1)*3^(n-1)+...18+32Tn=3Tn-Tn=n*3^n-3^(n-1)-3^(n-2)-....-3+1=1+n*3^n-[(1-3^n)/(1-3)]=1+n*3^n+1/2 -3^n/2=3/2+(n-1/2)*3^nTn=3/4 +(n/2 -1/4)*3^n