已知公差大于零的等差数列{an}的前n项和Sn,且满足a3*a4=117,a2+a5=22

1由a3*a4=117,a2+a5=22得(a1+2d)*(a1+3d)=117(a1+d)+(a1+4d)=22解得d=4（已知d为正）,a1=1则an=1+4(n-1)=4n-32Sn=n(a1+(n-1)d/2)=n(1+(n-1)*4/2)=n(2n-1)而bn=Sn/(n+c)=n(2n-1)/(n+c)为等差数列.则c为0或者-1/2.已知...