Question

已知数列{an},{bn}满足bn=an+1-an,其中n=1,2,3,… 若bn+1*bn-1=bn,且b1=1,b2

Answer 1

c(n) = a(6n-1),

把n+1当作n代入上式,有,

c(n+1) = a[6(n+1)-1] = a(6n+5),

所以,有,

c(n+1)-c(n) = a(6n+5) - a(6n-1)

= [a(6n+5) - a(6n+4)] + [a(6n+4)-a(6n+3)] + [a(6n+3)-a(6n+2)] + [a(6n+2)-a(6n+1)] + [a(6n+1)-a(6n)] + [a(6n) - a(6n-1)]

= b(6n+4) + b(6n+3) + b(6n+2) + b(6n+1) + b(6n) + b(6n-1)

...

这样,是否清楚一点了.