问题标题:
【Sn=a1n+[n{n-1}d]/2公式能否变成Sn=a2n+[n{n-2}d]/2】
问题描述:
Sn=a1n+[n{n-1}d]/2公式能否变成Sn=a2n+[n{n-2}d]/2
藏斌宇回答:
Sn=a1n+[n{n-1}d]/2公式能否变成Sn=a2n+[n{n-2}d]/2不对Sn=a1n+[n{n-1}d]/2=(a2-d)n+[n{n-1}d]/2=a2*n+[n(n-1)d-2dn]/2=a2n+[n(n-1-2)d=a2n+[n(n-3)d]/2还可表示Sn=am*n+n(n-m-1)d/2
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