
设数列1,An=an+b数列2,Bn=cn+d任何等差数列都可以化简成这样。数列Cn=An*Bn,现在就是求Cn的前n项和。An*Bn=(an+b)*(cn+d)=ac*n²+(ad+bc)*n+bdSn=ac(1²+2²+3²+。。n²)+(ad+bc)*(1+2+3+。。。n)+bd*n数列1²+2²+3²+。。n²=n(n+1)(2n+1)/6数列1+2+3+。。。n=n(n+1)/2则Sn=ac*n(n+1)(2n+1)/6+(ad+bc)*n(n+1)/2+bd*n

设数列1,An=an+b数列2,Bn=cn+d任何等差数列都可以化简成这样。数列Cn=An*Bn,现在就是求Cn的前n项和。An*Bn=(an+b)*(cn+d)=ac*n²+(ad+bc)*n+bdSn=ac(1²+2²+3²+。。n²)+(ad+bc)*(1+2+3+。。。n)+bd*n数列1²+2²+3²+。。n²=n(n+1)(2n+1)/6数列1+2+3+。。。n=n(n+1)/2则Sn=ac*n(n+1)(2n+1)/6+(ad+bc)*n(n+1)/2+bd*n