在数列{An}中，A1=1,An+1=(1+1/n)An+(n+1)/2^n.求数列的前N项和

Question

谢谢

Answer 1

∵An+1=[﹙n+1﹚/n]×An+(n+1)2^n ∴An+1/(n+1)=An/n+2^n 又设Bn=An/n① 得Bn+1=Bn+2^n② 设Bn+1+α×2^(n+1)=Bn+α×2^n ∴Bn+1=Bn+α×2^n③ ∴由②③知α=1 ∴Bn+1+2^(n+1)=Bn+2^n=Bn-1+2^(n-1)=…=B1+2 又∵B1=A1/1=1 ∴Bn+2^n=B1+2=3 ∴Bn=3-2^n ∴由①得An=n×Bn=n×(3-2^n)=3n-n×(2^n) ∴数列﹛An﹜的前n项和Sn=A1+A2+A3+…+An=3×1－1×2＋3×2－2×2^2＋…＋3×n－n×﹙2^n﹚ =3×﹙1＋2＋3＋…＋n﹚－﹙1×2＋2×2^2＋…＋n×2^n﹚ =3×﹙1＋n﹚×n／2－﹙1×2＋2×2^2＋…＋n×2^n﹚ 设Tn=1×2＋2×2^2＋…＋n×2^n④ ∴2Tn=0+1×2^2＋…＋﹙n－1﹚×2^n＋n×2^﹙n＋1﹚⑤ ∴由④－⑤得Tn－2n= - Tn=1×2＋2^2＋2^3＋…＋2^n－n×2^﹙n+1﹚ ∴Tn = - 2×﹙1－2^n﹚/﹙1－2﹚＋n×2^﹙n＋1﹚= - 2＋﹙n＋1﹚×2^﹙n＋1﹚ ∴Sn=3×﹙n＋1﹚×n／2－[ - 2＋﹙n＋1﹚×2^﹙n＋1﹚]=2＋3×n×﹙n＋1﹚／2－﹙n＋1﹚×2^﹙n＋1﹚