Question

数学题 关于数列

Answer 1

应该是这样

Answer 2

。

Answer 3

解：(1)∵a(n+1)=(1+1/n)an+(n+1)/2^n
=[(n+1)/n]an+(n+1)/2^n
∴a(n+1)/(n+1)=an/n+2^(-n)
又bn=an/n
∴b(n+1)=bn+2^(-n)
则b(n+1)-bn=2^(-n)
∴bn-b(n-1)=2^[-(n-1)]
… … …
b3-b2=2^(-2)
b2-b1=2^(-1)
以上等式相加，得
bn-b1=2^[-(n-1)]+2^[-(n-2)]+……+2^(-2)+2^(-1)
=(1/2)[1-(1/2)^(n-1)]/(1-1/2)
=1-(1/2)^(n-1)
故bn=b1+1-(1/2)^(n-1)
当n=1时，b1=a1/1=1
∴bn=2-(1/2)^(n-1)

追答

(2)由(1)，知bn=2-(1/2)^(n-1)
∴an=nbn=2n-n(1/2)^(n-1)
∴Sn=a1+a2+a3+……+an
=[2-(1/2)^0]+[2×2-2×(1/2)¹]+[2×3-3×(1/2)²]+……+[2n-n(1/2)^(n-1)]
=(2+4+6+……+2n)-[1+2×(1/2)+3×(1/2)²+……+n(1/2)^(n-1)]
=n(n+1)-[1+2×(1/2)+3×(1/2)²+……+n(1/2)^(n-1)]
令Tn=1+2×(1/2)+3×(1/2)²+……+n(1/2)^(n-1) ①
则(1/2)Tn=1/2+2×(1/2)²+3×(1/2)³+……+(n-1)(1/2)^(n-1)+n(1/2)^n ②
由①-②，得
(1/2)Tn=1+(1/2)+(1/2)²+(1/2)³+……+(1/2)^(n-1)-n(1/2)^n
=[1-(1/2)^n]/(1-1/2)-n(1/2)^n
=2-(1/2)^(n-1)-n(1/2)^n
=2-(2+n)(1/2)^n
∴Tn=4-(2+n)(1/2)^(n-1)
∴Sn=n(n+1)-[4-(2+n)(1/2)^(n-1)]
=n(n+1)-4+(2+n)(1/2)^(n-1)

Answer 4