Question

cho: S= \(3^0+3^1+3^2+...+3^{99}\)

CMR: S chia hết cho 4

Answer 1

S = 1 + 3 + 3² + ... + 3⁹⁹

   = ( 1 + 3 ) + ( 3² + 3³ ) + ... + ( 3⁹⁸ + 3⁹⁹ )

   = 1.4 + 3²(1+3) + ... + 3⁹⁸(1+3)

   = 4.(1+3²+...+3⁹⁸) chia hết cho 4

Answer 2

=> S = 1 + 3¹ + 3² + ........ + 3⁹⁹

= ( 1 + 3¹ ) + ( 3² + 3³ ) + .......... + ( 3⁹⁸ + 3⁹⁹ )

= 4 + 3²( 1 + 3¹ ) + ......... + 3⁹⁸( 1 + 3¹ )

= 4 . 3² . 4 + .......... + 3⁹⁸ . 4

= 4( 1 + ............ + 3⁹⁸ ) chia hết cho 4

=> ĐPCM

Answer 3

S=3⁰+3¹+3²+...+3⁹⁹

S=(3⁰+3¹)+(3²+3³)+...+(3⁹⁸+3⁹⁹)

S=3⁰.(1+3)+3².(1+3)+...+3⁹⁸.(1+3)

S=1.4+3².4+...3⁹⁸.4

S=4.(1+3²+...+3⁹⁸)

=>S chia hết cho 4