数列:2,3,4,6,8,12,16,24,32...的通项公式
问题描述:
数列:2,3,4,6,8,12,16,24,32...的通项公式
用二二叉法解,
答
设{an}:2,3,4,6,8,12,16,24,32,…,则{bn}={a(n+1)-an}:1,1,2,2,4,4,8,8,…,{cn}={b(n+1)-bn}:0,1,0,1,0,1,0,….于是b(n+1)-bn=cn=[1+(-1)^n]/2,运用迭加法,得bn=b1+c1+c2+…+c(n-1)=1+(n-1)/2 - [1+(-1)^n]/4= (2...