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意大利数学家列昂那多·斐波那契以兔子繁殖为例,引入“兔子数列”: \(1,1,2,3,5,8,13,21,34,…,\)即 \(F(1) = F(2) = 1,F(n) = F(n - 1) + F(n - 2)\left( {n⩾3,n \in {N^*}} \right)\) ,此数列在物理、化学等领域都有广泛的应用,若此数列被2整除后的余数构成一个新数列 \(\left\{ {{a_n}} \right\}\) ,则数列 \(\left\{ {{a_n}} \right\}\) 的前 2020 项的和为( )
A.1347
B.1348
C.1349
D.1346
参考答案:A
解析:
由数列\(1,1,2,3,5,8,13,21,34,…,\)各项除以\(2\)的余数,可得数列 \(\left\{ {{a_n}} \right\}\) 为\(1,1,0,1,1,0,1,1,0,…,\)所以数列\(\left\{ {{a_n}} \right\}\)是周期为3的周期数列,前三项和为 \(1 + 1 + 0 = 2\) ,又因为 \(2020 = 673 \times 3 + 1\) ,所以数列 \(\left\{ {{a_n}} \right\}\) 的前 2020 项的和为 \(673 \times 2 + 1 = 1347\) .
故选:A