If 'n' is a natural number, then (6n2+6n) is always divisible by-

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6 only - To determine what $6n^2 + 6n$ is always divisible by, we can factorize the expression and analyze its divisibility properties.1. Factor the Expression:  Given:   $$  6n^2 + 6n  $$  Factor out the common factor of 6:  $$  6n^2 + 6n = 6(n^2 + n)  $$2. Factorize $n^2 + n$:  The expression inside the parentheses can be factored further:  $$  n^2 + n = n(n + 1)  $$  Therefore:  $$  6(n^2 + n) = 6n(n + 1)  $$3. Analyze Divisibility:  - $n$ and $n + 1$ are consecutive integers.   - Consecutive integers are always such that one of them is even, which means $n(n + 1)$ is always divisible by 2.  Therefore, $6n(n + 1)$ is always divisible by:  - 6 (from the factor outside the parentheses).  - 2 (from the fact that one of $n$ or $n + 1$ is even).4. Check Divisibility by 3:  - Among any two consecutive integers, one is divisible by 3. Thus, $n(n + 1)$ is always divisible by 3.  Since $6 = 2 \times 3$, the expression $6n(n + 1)$ is divisible by:  - 2  - 3  Combining these, $6n(n + 1)$ is always divisible by:  - 65. Verify:  Let's check a few values to confirm:  - For $n = 1$: $6n^2 + 6n = 6(1 + 1) = 12$ (divisible by 6)  - For $n = 2$: $6n^2 + 6n = 6(4 + 2) = 36$ (divisible by 6)  - For $n = 3$: $6n^2 + 6n = 6(9 + 3) = 72$ (divisible by 6)So, the expression $6n^2 + 6n$ is always divisible by $6$.

If 'n' is a natural number, then (6n2+6n) is always divisible by-

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