If n is a natural number, then (6n2 + 6n) is always divisible by:Options :Both 6 & 12612none

The given expression is 6n^2 + 6n.

Step 1: Factor out a 6 from the expression. This gives us 6(n^2 + n).

Step 2: Notice that n^2 + n is always a natural number because it is the sum of two natural numbers.

Step 3: Since 6 times any natural number is always divisible by 6, we can say that 6n^2 + 6n is always divisible by 6.

Step 4: To check if the expression is divisible by 12, we need to see if n^2 + n is always even. This is because 6 times an even number is always divisible by 12.

Step 5: If n is even, then n^2 + n is even because the sum of two even numbers is always even. If n is odd, then n^2 is odd and n is odd, and the sum of two odd numbers is always even. So, n^2 + n is always even.

Step 6: Therefore, 6n^2 + 6n is always divisible by 12.

So, the answer is that (6n^2 + 6n) is always divisible by both 6 and 12.