a2-a 1= 1，a3+a2=3，a4-a3=5，a5+a4=7，a6-a5=9，a7+a6= 1 1，…a50-a49=97，

The deformation {an} can be obtained as a3+a 1=2, a4+a2=8, a7+a5=2, a8+a6=24, a9+a7=2, a12+a/kloc-0 = 40, a.

Solution: Solution: Because the sequence {an} satisfies an+1+(-1) n an = 2n-1,there are a2-a 1= 1, a3+a2=3, A4-A3.

So we can get a3+a 1=2, a4+a2=8, a7+a5=2, a8+a6=24, a9+a7=2, a 12+a 10=40, A/kloc-.

Starting from the first term, the sum of two adjacent odd-numbered terms is equal to 2, and starting from the second term, the sum of two adjacent even-numbered terms is taken in turn to form the arithmetic progression of the first term with 8 bits and a tolerance of 16.

{an} The sum of the first 60 items is15× 2+15× 8+(15×14) ×16/2) =1830.

Comments: This question mainly examines the method of summation of series, arithmetic progression's summation formula, and pays attention to the structural characteristics of series, which is an intermediate question.

I hope it helps you.

(Note that the following A is also a subscript here.