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
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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n项和Sn
麻烦说清楚点,
麻烦说清楚点,
an=a1+(n-1)(2/3)=(2n+1)/3
bn=(-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]
bn=(-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]
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