Z,
such that 4k = 6. 6. To disprove this statement, you need only find ONE value of n (an even integer) for which the statement does not hold.
Suppose n = 4.
| Then the left hand side of the equation becomes: | 1 + 2 + 3 = 6, |
| and the right hand side of the equation becomes: | 4k. |
There is no value of k Z,
such that 4k = 6.