1、
an=1+(n-1)d;
bn = q^(n-1);
a2 = 1+d = b2 = q;
a5 = 1+4d = q^2 = (1+d)^2;
得出d(d-2)=0;
因为a2=b2 ≠1,所以d=2,q = 3；
an = 2n-1;bn = 3^(n-1)
所以 cn = (2n-1)*3^(n-1);
2、
Sn = 1^3^0+3*3^1+5*3^2+.+(2n-1)*3^(n-1)
3Sn = 1*3^1+3*3^2+.+(2n-3)*3^(n-1)+(2n-1)*3^n
3Sn-Sn = -1-2*3^1 - 2*3^2 -.-2*3^(n-1)+(2n-1)*3^n
=-1-2[(3^n-1)/2 -1]+(2n-1)*3^n
=-1-3^n+1+2+(2n-1)*3^n
=(2n-2)*3^n+2
Sn = (n-1)*3^n+1