数列an中,a1=0.5,an+1=an+an^2,bn=1/(1+an),sn=b1+b2=b3……+bn,pn=b1b2b3……bn,求2pn+sn的值

Question

数列an中,a1=0.5,an+1=an+an^2,bn=1/(1+an),sn=b1+b2=b3……+bn,pn=b1b2b3……bn,求2pn+sn的值

Answer 1

a(n+1)=an(1+an)
所以bn=1/(1+an)=an/a(n+1)
又由a(n+1)=an+an^2得（两边同除ana(n+1)）
1/an=1/a(n+1)+an/a(n+1))=1/a(n+1)+bn
所以bn=1/an-1/a(n+1)
所以sn=b1+b2+...+bn=1/a1-1/a2+1/a2-1/a3+...+1/an-1/a(n+1)
=1/a1-1/a(n+1)
pn=b1b2...bn=(a1/a2)(a2/a3)...(an/a(n+1))
=a1/a(n+1)
所以2pn=2a1/a(n+1)=1/a(n+1)
所以sn+2pn=1/a1-1/a(n+1)+1/a(n+1)
=1/a1=2