Question

求{bn}的通项公式,急,a1=1,b1=1任意的n∈n+,均有a=[(anf)/(f+3)],b-bn=1/an求bn

Answer 1

-bn=1/an=3n-2

bn-b(n-1)=3(n-1)-2

.

b3-b2=3x2-1

b2-b1=3x1-1

以上各式相加得：bn-b(n-1)+b(n-1)-b(n-2)+...+b3-b2+b2-b1=

3(1+2+3+...+n-1)-2(n-1)

bn-b1=3n(n-1)/2-2n+2

bn=3-(3n-n²)/2,且b1=1