已知数列{an}的通项公式an=cos n派
已知数列{an}的通项公式an=cos n派/12,则该数列的前12项和为求详解!
该数列的前12项和=∑cos (nπ/12) cos(nπ/12)=-cos(π-nπ/12) cos(π/12)=-cos(11π/12) cos(2π/12)=-cos(10π/12) cos(3π/12)=-cos(9π/12) cos(4π/12)=-cos(8π/12) cos(5π/12)=-cos(7π/12) 该数列的前12项和=∑cos (nπ/12)=cos(6π/12)+cos(12π/12)=0+(-1) =-1
答:an=1/n(n+1)=(1/n)-(1/n+1) a(n-1)=(1/n-1)-(1/n) a(n-2)=(1/n-2)-(1/n-1) ...... a(1...详情>>