等差数列SN=20 S2N=38,求S3N
问题描述：
等差数列SN=20 S2N=38,求S3N
答
Sn=na1+n（n-1)d/2=na1+(n^2d/2)-(nd/2)S2n-Sn=na1+(3n^2 d/2) -(nd/2)S3n-S2n=na1+(5n^2 d/2)-(nd/2)发现2（S2n-Sn）=(S3n-S2n )+Sn所以Sn,S2n-Sn,S3n-S2n 成等差数列依题意,S3n-S2n=16所以S3n=54就是基本量法...