数列{an}的通项公式an=n·cos(n·π)⼀2 +1,前n项和为Sn,求S2012

2025-06-27 09:24:08
推荐回答(2个)
回答1:

a1=cosπ/2+1
a2=2cos2π/2+1
a3=3cos3π/2+1
a4=4cos4π/2+1
...
Sn = S{ncosnπ}/2 + n
S{ncosnπ} = -1+2-3+4-5+6+...+n*(-1)^n
S2012 = (2012/2 )/2+2012 = 2515

回答2:

3016