第1个回答 2017-06-17
通项 an=a1+(n-1)d
过程:a1=a1+0d=a1+(1-1)d
a2=a1+1d=a1+(2-1)d
a3=a2+d=a1+2d=a1+(3-1)d
a4=a3+d=a1+3d=a1+(4-1)d
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an=a1+(n-1)d。
前n项和Sn=n(a1+an)/2
过程:Sn=a1+a2+a3+……+a(n-1)+an
Sn=an+a(n-1)+a(n-2)+……+a2+a1
两式相加
2Sn=(a1+an)+(a2+an-1)+(a3+an-2)+……+(an-1+a2)+(an+a1)=(an+a1)+(an+a1)+(an+a1)+……+(an+a1)+(an+a1)
一共n项(n+1)
2Sn=n(an+a1)
Sn=n(an+a1)/2