如题所述
解:通项是an=n²求前n项和Sn因为(n+1)³-n³=3n²+3n+12³-1³=3*1²+3*1+13³-2³=3*2²+3*1+1......n³-(n-1)³=3(n-1)²+3(n-1)+1(n+1)³-n³=3n²+3n+1累加得;(n+1)³-1=3Sn+3(1+2+...+n)+n(n+1)³-1=3Sn+3n(n+1)/2+n所以Sn=n(n+1)(2n+1)/6