Question

（2012•黑龍江）數(shù)列{a_n}滿足an+1+(-1)nan=2n-1，則{a_n}的前60項(xiàng)和為

1830

．

Answer 1

分析：令b_n+1=a_4n+1+a_4n+2+a_4n+3+a_4n+4，則b_n+1=a_4n+1+a_4n+2+a_4n+3+a_4n+4=a_4n-3+a_4n-2+a_4n-2+a_4n+16=b_n+16可得數(shù)列{b_n}是以16為公差的等差數(shù)列，而{a_n}的前60項(xiàng)和為即為數(shù)列{b_n}的前15項(xiàng)和，由等差數(shù)列的求和公式可求

解答：解：∵an+1+(-1)nan=2n-1，
∴an+1=2n-1-(-1)nan
令b_n+1=a_4n+1+a_4n+2+a_4n+3+a_4n+4
則b_n+1=a_4n+1+a_4n+2+a_4n+3+a_4n+4=a_4n-3+a_4n-2+a_4n-2+a_4n+16=b_n+16
∴數(shù)列{b_n}是以16為公差的等差數(shù)列，{a_n}的前60項(xiàng)和為即為數(shù)列{b_n}的前15項(xiàng)和
∵b₁=a₁+a₂+a₃+a₄=10
∴S=10×15+

15×14

2

×16=1830

點(diǎn)評(píng)：本題主要考查了由數(shù)列的遞推公式求解數(shù)列的和，等差數(shù)列的求和公式的應(yīng)用，解題的關(guān)鍵是通過(guò)構(gòu)造等差數(shù)列