N is natural number then 6n square + 6n is always divisible by ?
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6n^2+6n
6n(n+1)
so n+1 is a factor of this equation
and it's always divisible by its factor
6n(n+1)
so n+1 is a factor of this equation
and it's always divisible by its factor
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6n^2 + 6n = 6(n(n+1))
Clearly it is divisible by 6. As n(n+1) is always even, 6(n(n+1)) is always divisible by 12
Clearly it is divisible by 6. As n(n+1) is always even, 6(n(n+1)) is always divisible by 12
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