已知等差数列｛an｝的首项a1=1,公差d=2/3,且bn=（-1的n-1次方）乘以an乘以an+1,求数列｛bn｝的前

an=a1+(n-1)(2/3)=(2n+1)/3bn=(-1)^(n-1)*an*a(n+1)当n是偶数时Sn=b1+b2+。。。+bn=a1a2-a2a3+a3a4-a4a5+。-ana(n+1)=a2(a1-a3)+a4(a3-a5)+。+an[a(n-1)-a(n+1)]=(a2+a4+a6+。+an)*(-2d)=[(a2+an)/2]*(n/2)*(-2d)=-[5/3+(2n+1)/3]*(n/3)=-2n(n+3)/9当n是奇数时Sn=S(n-1)+bn=-2(n-1)(n+2)/9 + [(2n+1)/3][(2n+3)/3]