（Ⅰ）∵m→∥n→∴1x3+c-1•y-1=0⇒y=x3+c-1(x3+c-1≠0),因?yàn)楹瘮?shù)f（x）為奇函數(shù)．所以c=1,⇒f（x）=x3（x≠0）（Ⅱ）由題意可知,f（a1）+f（a2）+…+f（an）=Sn2⇒a13+a23+a33+…+an3=Sn2…..①n≥2時(shí)∴a13+a23+a33+…+an-13=Sn-12…②由①-②可得：an3=Sn2-Sn-12=an（Sn+Sn-1）,∵{an}為正數(shù)數(shù)列∴an2=Sn+Sn-1…③∴an+12=Sn+1+Sn…④由④-③可得：an+12-an2=an+1+an∵an+1+an＞0,∴an+1-an=1,且由①可得a13=a12,a1＞0⇒a1=1,a13+a23=S22,a2＞0⇒a2=2,∴a2-a1=1∴{an}為公差為1的等差數(shù)列,∴an=n（n∈N*）（Ⅲ）∵an=n（n∈N*）,∴bn=4n-a•2n+1=（2n-a）2-a2（n∈N*）令2n=t（t≥2）,∴bn=（t-a）2-a2（t≥2）（1）當(dāng)a≤2時(shí),數(shù)列{bn}的最小值為：當(dāng)n=1時(shí),b1=4-4a．（2）當(dāng)a＞2時(shí)①若a=2k+1（k∈N*）時(shí),數(shù)列{bn}的最小值為當(dāng)n=k+1時(shí),bk+1=-a2．②若a=2k+2k+12(k∈N*)時(shí),數(shù)列{bn}的最小值為,當(dāng)n=k或n=k+1時(shí),bk=bk+1=（2k-a）2-a2．③若2k＜a＜2k+2k+12(k∈N*)時(shí),數(shù)列{bn}的最小值為,當(dāng)n=k時(shí),bk=（2k-a）2-a2④若2k+2k+12＜a＜2k+1(k∈N*)時(shí),數(shù)列{bn}的最小值為,當(dāng)n=k+1時(shí),bk+1=（2k+1-a）2-a2．