Question

Cho B = 3^1 + 3^2 + 3^3 + ..... + 3^100 . Chứng minh b chia hết cho 2

Answer 1

\(B=3\left(1+3\right)+...+3^{99}\left(1+3\right)\)

\(=4\cdot\left(3+...+3^{99}\right)⋮2\)

Answer 2

\(B = 3 1 + 3 2 + 3 3 + . . . . . + 3 100 = ( 3 + 3 2 ) + ( 3 3 + 3 4 ) + . . . + ( 3 99 + 3 100 ) = 3 ( 1 + 3 ) + 3 3 ( 1 + 3 ) + . . . + 3 99 ( 1 + 3 ) = 3.4 + 3 3 .4 + . . . + 3 99 .4 = 4 ( 3 + 3 3 + . . . + 3 99 ) d o : 4 ⋮ 2 => 4 ( 3 + 3 3 + . . . + 3 99 ) ⋮ 2 => B ⋮ 2 vậy B chia hết cho 2\)