已知数列{an}是等差数列,且a1=2,a1+a2+a3=12,求数列{an}的通项公式;设bn=

2025-05-13 18:35:17
推荐回答(3个)
回答(1):

a2 = a1 + d; a3 = a1 + 2d;
则 a1+a2+a3 = 3 得出d = 2;
所以an = a1 + (n - 1)d = 2 + (n - 1)*2 = 2n

回答(2):

An=2n

回答(3):