Question

Tìm n thuộc Z .

a ) ( n + 1 ) + ( n +3 ) + ( n + 5 ) + ... + ( n + 99 ) = 0

b ) ( n - 3 ) + ( n - 2 ) + ( n - 1 ) + ... + 10 + 11 = 11

Answer 1

a) ta có ( n + 1 ) + ( n + 3 ) + ( n + 5 ) + ....+ ( n + 99 ) = 0

=> (n + n + n +......+ n )  + ( 1 + 3 + 5 +.....+ 99 ) =0

=> ( n x 50 ) + [ ( 99 - 1 ) x ( 99 + 1 ) ] : 2 = 0

=> ( n x 50 ) + 2500 = 0 

=> ( n x 50 ) = 0 - 2500

=> n x 50 = -2500

=> n = 2-2500 : 50 = -50

Answer 2

a) ta có ( n + 1 ) + ( n + 3 ) + ( n + 5 ) + ....+ ( n + 99 ) = 0

=> (n + n + n +......+ n )  + ( 1 + 3 + 5 +.....+ 99 ) =0

=> ( n x 50 ) + [ ( 99 - 1 ) x ( 99 + 1 ) ] : 2 = 0

=> ( n x 50 ) + 2500 = 0 

=> ( n x 50 ) = 0 - 2500

=> n x 50 = -2500

=> n = 2-2500 : 50 = -50

phần b tự làm nha

Answer 3

lol oki

faker oki 

SKTT 1oki