Question

数列{an}满足an+1+（-1）nan=2n-1，则{an}的前60项和为______

数列{an}满足an+1+（-1）nan=2n-1，则{an}的前60项和为______．

Answer 1

∵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
则b_n+1=a_4n+1+a_4n+2+a_4n+3+a_4n+4=a_4n-3+a_4n-2+a_4n-1+a_4n+16=b_n+16
∴数列{b_n}是以16为公差的等差数列，{a_n}的前60项和为即为数列{b_n}的前15项和
∵b₁=a₁+a₂+a₃+a₄=10
∴S＝10×15+

15×14

2

×16=1830