a, 1+3+3^2+3^3+...+3^48+3^49
=(1+3)+(3^2+3^3)+...+(3^48+3^49)
=(1+3)*(1+3^2+3^2+...+3^48+3^48)
=4*(1+3^2+3^2+...+3^48+3^48) chia hết cho 4 (đccm)
b,
S=1+3+3^2+3^3+...+3^48+3^49
3S=3+3^2+3^3+3^4+...+3^49+3^50
3S-S=3^50-1
2S=2^50
S=2^50 : 2
S=1^50 = 1
vậy tận cùng của S =1
c,
93^50-1)/2
=2^50/2
=1^50
=1