4B = 1.2.3.4+2.3.4.4+....+(n-1).n.(n+1).4
= 1.2.3.4+2.3.4.(5-1)+....+(n-1).n.(n+1).[(n+2)-(n-2)]
= 1.2.3.4+2.3.4.5-1.2.3.4+....+(n-1).n.(n+1).(n+2)-(n-2).(n-1).n.(n+1)
= (n-1).n.(n+1).(n+2)
=> B = (n-1).n.(n+1).(n+2)/4
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