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
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
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