题目

数列an的通项公式an=6n-5 k=2n-1 k属于n* an=4的n次方 n=2k k属于n*
求an的前2n项的和

提问时间：2021-03-21

答案

an =6n-5 ; n is odd number
= 4^n ; n is even number
for n is odd
an = 6n-5
a1+a3+a5+...+a(2n-1) = (12n -10)n (1)
for n is even
a2+a4+...+a2n = 8(4^(2n) -1) /(4-1) = (8/3)(4^(2n) -1 ) (2)
(1)+(2)
a1+a2+...+a(2n) =(12n -10)n +(8/3)(4^(2n) -1 )