等差数列an前n项和为Sn,求证S(2n-1)=(2n-1)an
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/21 13:00:38
![等差数列an前n项和为Sn,求证S(2n-1)=(2n-1)an](/uploads/image/z/8407622-38-2.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%B1%82%E8%AF%81S%EF%BC%882n-1%EF%BC%89%3D%EF%BC%882n-1%EF%BC%89an)
等差数列an前n项和为Sn,求证S(2n-1)=(2n-1)an
等差数列an前n项和为Sn,求证S(2n-1)=(2n-1)an
等差数列an前n项和为Sn,求证S(2n-1)=(2n-1)an
S(2n-1)=(a1+a2n-1)(2n-1)/2=(2n-1)[a1+a1+(2n-2)d]/2=(2n-1)(a1+(n-1)d)=(2n-1)an