Ta có: S=1+3+3^2+3^3+...+3^99
S = (1+3^1+3^2+3^3) + (3^5+3^6+3^7+3^8) + ... + (3^96+3^97+3^98+3^99) (cứ 4 số hạng gộp lại)
S=(1+3^1+3^2+3^3) + 3^5(1+3^1+3^2+3^3) + ...+3^96(1+3^1+3^2+3^3)
Mà 1+3^1+3^2+3^3 = 40
Nên S= 40 + 3^5.40 +... + 3^96.40
=40.(1+3^5+...+3^96)
=10.4(1+3^5+..+3^96) ( chia hết cho 10)
Vậy S chia hết cho 10
S= ( 1+3+3^2))+...+(3^98+3^99)
=3*(1+3^2)+..+3^98*(1+3^2)
=3*4+...+3^98*4
=3*4+...+3^99*3*4
=12+...+3^99*12
=S=(1+...+3^99)*10 chia het cho10
=> S chia het cho 10
Minh nghi la vayt vi minh cung ko chac la dung neu sai thi mong ban thong cam !