Question

數(shù)列{a_n}滿足a1=a2=1，an+an+1+an+2=cos

2nπ

3

(n∈N*)，數(shù)列前n項(xiàng)和為S_n，則S₂₀₁₃=

-

671

2

．

Answer 1

分析：當(dāng)n=3k-2時(shí)（k∈N^*），a_3k-2+a_3k-1+a_3k=cos

2(3k-2)π

3

=-

1

2

．即可得出S₂₀₁₃=（a₁+a₂+a₃）+…+（a₂₀₁₁+a₂₀₁₂+a₂₀₁₃）．

解答：解：當(dāng)n=3k-2時(shí)（k∈N^*），a_3k-2+a_3k-1+a_3k=cos

2(3k-2)π

3

=cos(2kπ-

4π

3

)=-cos

π

3

=-

1

2

．
∴S₂₀₁₃=（a₁+a₂+a₃）+…+（a₂₀₁₁+a₂₀₁₂+a₂₀₁₃）
=-

1

2

×671
=-

671

2

．
故答案為-

671

2

．

點(diǎn)評(píng)：本題考查了數(shù)列的周期性，屬于難題．