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

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:
  - 6

5. 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 \(6\).