Question

已知数列an的通项公式为an=（n+1）*（10/11）的n次幂.当项数n为多少时,an最大.

Answer 1

因为数列的每一项都不为负

那么

An+1 - An > 0

=> An+1 / An >1

=> [(n+2)*(10/11)^(n+1)] / [(n+1)*(10/11)^n] > 1

=> [n+2]/[n+1] > 11/10

=> n 0

n=9 => An+1 - An = 0,A10 = A9

n>9 => An+1 - An < 0

数列的最大项为A9和A10