If 'n' is a natural number, then (6n2+6n) is always divisible by-
Here, 6n2 + 6n = 6(n2 + n) which is always divisible by 6.
as n is a natural number so, 6 ( n2 + n) for values of n= 1,2,3,4,5, ………. are
for 1 it will be = 6 ( 12 + 1) = 12
for 2 it will be = 6 ( 22 + 2) = 36
for 3 it will be = 6 ( 32 + 3) = 72
for 4 it will be = 6 ( 42 + 4) = 120
for 5 it will be = 6 ( 52 + 5) = 180 ….
as we can see the result will always be divisible by both 6 and 12. And it happens for the square of n.
in other terms, we can say, it is = 6n (n+1)
as n is natural so it can not be less than 1 so from the least to each values of n we can see that 6n is always multiplied by minimum (n +1) = (1+1) = 2 or its multiplicative.
so the result will be at least 12n or its multiplicative.
so it will always be divisible by both 6 and 12.
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