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{an}等差數列,其公差為d則 Sn=a1+(a1+d)+(a1+2d)+…+[a1+(n-1)d], Sn=an+(an-d)+(an-2d)+…+[an-(n-1)d], 兩式相加得2Sn=n(a1+an)∴Sn=n(a1+an)/2
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