已知等差数列{an}的前N项和为Sn,a1=-2/3,满足Sn+1/Sn+2=an(n大于等于2) 求Sn 别用数学归纳
3个回答
Sn+1/Sn+2=an带入n=2 又an=a1+(n-1)d 故得(3a1+2d)/(4a1+6d)=a1+d
带入a1 即得d Sn=na1+n(n-1)d/2
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