等差数列{an}前n项和为sn,求证S2n-1=(2n-1)an
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等差数列{an}前n项和为sn,求证S2n-1=(2n-1)an等差数列{an}前n项和为sn,求证S2n-1=(2n-1)an等差数列{an}前n项和为sn,求证S2n-1=(2n-1)an我先和你
等差数列{an}前n项和为sn,求证S2n-1=(2n-1)an
等差数列{an}前n项和为sn,求证S2n-1=(2n-1)an
等差数列{an}前n项和为sn,求证S2n-1=(2n-1)an
我先和你讲一个直白一点的
S3=3*a2
S11=11*a6
一般来说,奇数个等差数列的和等于个数乘以这个数列的中位数
所以S2n-1中,其中位数为an
所以S2n-1=(2n-1)an
望采纳
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