Question

如何构造累差迭加特殊数列在数列an中,a1=1,an+1=(1+1/n)an+(n+1)/2的n次方（I）设bn=an/

Answer 1

a(n+1)=(1+1/n)an+(n+1)/2^n

bn=an/n,an=nbn

(n+1)b(n+1)=(1+1/n)nbn+(n+1)/2^n

b(n+1)=bn+1/2^n=bn+（1/2)^n

bn=b(n-1)+)(1/2)^(n-1)

.

bz=b1+(1/2)^1

相加的：bn=b1+(1/2)^(1)+.+(1/2)^(n-1),b1=a1/1=1

bn=1+(1/2)^(1)+.+(1/2)^(n-1)=(1/2)^0+(1/2)^(1)+.+(1/2)^(n-1)

bn=(1-(1/2)^n)/(1-1/2)=2(1-(1/2)^n)=2(1-1/(2)^n)=2-1/2^(n-1)

an=2n-n/2^(n-1),