第1个回答 2011-09-27
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﹚
第2个回答 2011-09-27
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/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+1/2-1/2^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)