Question

已知数列an的首项a1=3R,对任意自然数n都有2R/（an-an+1）=n(n+1)

Answer 1

第二问应该是bn=R^n/(a1a2a3……an)?

(1)

2R/（an-an+1）=n(n+1),an+1-an= -2R/n(n+1)= -2R[1/n-1/(n+1）],得到：

a2-a1= -2R(1-1/2),a3-a2= -2R(1/2-1/3).an-1-an-2= -2R[1/(n-2)-1/(n-1)],an-an-1=-2R[1/(n-1)-1/n],将以上等式进行左加左、右加右叠加得到：an-a1= -2R（1-1/n),将a1=3R代入得到an=R(1+2/n)=R(n+2)/n

数列an通项为an=R(1+2/n)=R(n+2)/n

(2)

a1a2.an=R(3/1)R(4/2)R(5/3).R(n+2)/n=R^n[(n+1)(n+2)/2]

bn=R^n/(a1a2a3……an)=2/[(n+1)(n+2)]=2[1/(n+1)-1/(n+2)]

Sbn=b1+b2+.+bn=2(1/2-1/3)+2(1/3-1/4)6+.2[1/(n+1)-1/(n+2)]

=2[1/2-1/(n+2)]

(n→﹢∞)limSbn=(n→﹢∞)lim2[1/2-1/(n+2)]=1

数列bn的前n项和为2[1/2-1/(n+2)],极限为1