若数列{an}满足an+1+(-1)n an =2n-1，则{an}的前60项和为 .
若数列{an}满足an+1+(-1)n an =2n-1，则{an}的前60项和为 .
1 830 【解析】当n=2k-1时，a2k-a2k-1=4k-3；当n=2k时，a2k+1+a2k=4k-1；当n=2k+1时，a2k+2-a2k+1=4k+1.将a2k+1+a2k=4k-1与a2k-a2k-1=4k-3相减得a2k+1+a2k-1=2；将a2k+1+a2k=4k-1与a2k+2-a2k+1=4k+1相加得a2k+2+a2k=8k.所以(a1+a3)+(a5+a7)+…+(a57+a59)=15×2=30，(a2+a4)+(a6+a8)+(a10+a12)+…+(a58+a60)=8+24+40+…+232=
=1 800，所以S60=30+1 800=1 830.