作业帮已知数列{An}是一个首项为1,公差为2/3的等差数列,Bn=[(-1)^(n-1)]*An*A(n+1),\x0d设数列{Bn}的前n项和为Sn,求Sn.\x0d答案是：\x0dSn=-2n^2/9-2n/3(n为偶数)\x0d 或=2n^2/9+2n/3+7/9(n为奇数)\x0d这是怎么做出来的?\x0d
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/05 13:03:13
![已知数列{An}是一个首项为1,公差为2/3的等差数列,Bn=[(-1)^(n-1)]*An*A(n+1),\x0d设数列{Bn}的前n项和为Sn,求Sn.\x0d答案是：\x0dSn=-2n^2/9-2n/3(n为偶数)\x0d 或=2n^2/9+2n/3+7/9(n为奇数)\x0d这是怎么做出来的?\x0d](/uploads/image/z/2782331-35-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7BAn%7D%E6%98%AF%E4%B8%80%E4%B8%AA%E9%A6%96%E9%A1%B9%E4%B8%BA1%2C%E5%85%AC%E5%B7%AE%E4%B8%BA2%2F3%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2CBn%3D%5B%28-1%29%5E%28n-1%29%5D%2AAn%2AA%28n%2B1%29%2C%5Cx0d%E8%AE%BE%E6%95%B0%E5%88%97%7BBn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%B1%82Sn.%5Cx0d%E7%AD%94%E6%A1%88%E6%98%AF%EF%BC%9A%5Cx0dSn%3D-2n%5E2%2F9-2n%2F3%28n%E4%B8%BA%E5%81%B6%E6%95%B0%29%5Cx0d+%E6%88%96%3D2n%5E2%2F9%2B2n%2F3%2B7%2F9%28n%E4%B8%BA%E5%A5%87%E6%95%B0%29%5Cx0d%E8%BF%99%E6%98%AF%E6%80%8E%E4%B9%88%E5%81%9A%E5%87%BA%E6%9D%A5%E7%9A%84%3F%5Cx0d)
已知数列{An}是一个首项为1,公差为2/3的等差数列,Bn=[(-1)^(n-1)]*An*A(n+1),\x0d设数列{Bn}的前n项和为Sn,求Sn.\x0d答案是：\x0dSn=-2n^2/9-2n/3(n为偶数)\x0d 或=2n^2/9+2n/3+7/9(n为奇数)\x0d这是怎么做出来的?\x0d
已知数列{An}是一个首项为1,公差为2/3的等差数列,Bn=[(-1)^(n-1)]*An*A(n+1),\x0d设数列{Bn}的前n项和为Sn,求Sn.\x0d答案是：\x0dSn=-2n^2/9-2n/3(n为偶数)\x0d 或=2n^2/9+2n/3+7/9(n为奇数)\x0d这是怎么做出来的?\x0d图见：\x0d
\x0d请写出详细过程及思路,
已知数列{An}是一个首项为1,公差为2/3的等差数列,Bn=[(-1)^(n-1)]*An*A(n+1),\x0d设数列{Bn}的前n项和为Sn,求Sn.\x0d答案是：\x0dSn=-2n^2/9-2n/3(n为偶数)\x0d 或=2n^2/9+2n/3+7/9(n为奇数)\x0d这是怎么做出来的?\x0d
本题考查的是数列重组后新数列的性质问题
当n=2k时,(相邻两项提公因式后,变成n/2个特殊数列公差为4/3)
Sn=b1+b2+...+b2k
=A1A2-A2A3+A3A4-A4A5+...+A(2k-1)A2k-A2kA(2k-1)
=A2(A1-A3)+A4(A3-A5)+...+A2k(A2k-1-A2k+1)
=-4/3(A2+A4+...A2k)
=2n^2/9-2n/3
当n=2k-1时(去掉第一项后,相邻两项提公因式后重组成k-1项的等差数列)
Sn=b1+b2+b3+...+b(2k-1)
=A1A2-A2A3+A3A4-A4A5+...-A(2k-2)A(2k-1)+A(2k-1)A2k
=A1A2-A3(A2-A4)-A5(A4-A6)-...A(2k-1)[A(2k-2)-A(2k)]
=1*(5/3)+4/3[A3+A5+..A(2k-1)]
=2n^2/9+2n/3+7/9