Question

在数列{an}中,a1=1,an+1=(1+1/n)an+(n+1)/2^n

(1)设bn=an/n,求数列{bn}的通项公式
(2)求数列{an}的前n项和Sn

Answer 1

(1) ∵贺隐an+1=(1+1/n)an+(n+1)/2^n=(n+1)/n*an+(n+1)/2^n
∴an+1/(n+1)=an/n+1/2^n
∵bn=an/n，∴bn+1=bn+1/2^n
bn=bn-1+1/2^(n-1)
bn-1=bn-2+1/2^(n-2)
......
b2=b1+1/2^1
b1=a1/1=1
将上述n个式子加起来，得
bn=1+1/2+1/2^2+...+1/2^(n-1)
=1+1/2(1-(1/2)^(n-1))/(1-1/2)
=1+1-1/2^(n-1)
=2-1/2^(n-1)
(2) ∵bn=an/n，∴an=n*bn
∴Sn=1*b1+2*b2+...+n*bn
=1*(2-1)+2*(2-1/2)+3*(2-1/4)+...+n*(2-1/2^(n-1))
=2(1+2+3+...+n)-(1+2/2+3/4+...+n/局拍山2^(n-1))
=2Pn-Qn
设Pn=(1+2+3+...+n)=n(n+1)/2 Qn=(1+2/2+3/2^2+4/2^3+...+n/2^(n-1))
对Qn，有 Qn/2=(1/2+2/2^2+3/2^3+...+(n-1)/2^(n-1)+n/2^n)
Qn-Qn/2=Qn/2=1+[1/2+1/2^2+1/2^3+...+1/2^(n-1)]+n/2^n
=1+1/桐中2(1-(1/2)^(n-1))/(1-1/2)+n/2^n
=1+1-(1/2)^(n-1)+n/2^n
=2-1/2^(n-1)+n/2^n
=2+(n+2)/2^n
∴Qn=4+2(n+2)/2^n
∴Sn=2Pn-Qn
=2*n(n+1)/2-4-2(n+2)/2^n
=n(n+1)-4-2(n+2)/2^n

希望对你有帮助

Answer 2

由an+1=(1+1/n)an+(n+1)/2^n得an+1=(n+1)an/n+(n+1)/2^n,也就是an+1/(n+1)-an/n=1/2^n

a2/2-a1=1/2
a3/3-a2/2=1/2^2
a4/4-a3/3=1/2^3
......
an/n--an-1/(n-1)=1/2^(n-1)
把上面n个式子相加抵消中间项得an/n-a1=1/2+...+1/2^(n-1)
=1/2(1-1/2^(n-1))/(1-1/2)=1-1/2^(n-1)
所以an/n=a1+1-1/2^(n-1) =2-1/2^(n-1),{an}通项公式为an=n[2-1/2^(n-1)].
bn=2-1/2^(n-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知厅迹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－﹙喊伏锋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－﹙郑晌n＋1﹚×2^﹙n＋1﹚

Answer 3

a(n+1)=(1+1/n)a(n)+(n+1)/2^n=(n+1)a(n)/n+(n+1)/2^n
so a(n+1)/(n+1)=a(n)/n+1/2^n
so a(n+1)2^(n+1)/(n+1)=2(a(n)2^n/兆含凯n)+2
so a(n+1)2^(n+1)/(n+1)+2=2(a(n)2^n/n+2)
assume c(n)=a(n)2^n/n+2
then c(n+1)=2c(n) and c(1)=1X2/1+2=4
so c(n)=4X2^(n-1)
so a(n)=(4X2^(n-1)-2)n/2^n=n(2-2^(1-n))
so b(n)=a(n)/n=2-2^(1-n)
so S(n)=2X(1+n)n/2-2(1-1/2^n+1/族唤老此2-1/2^n+...+1/2^(n-1)-1/2^n)
=n(n+1)-2(1+1-1/2^(n-1)-n/2^n)
=n(n+1)-4+(2+n)/2^(n-1)