Base case: if <em>n</em> = 1, then
1² - 1 = 0
which is even.
Induction hypothesis: assume the statement is true for <em>n</em> = <em>k</em>, namely that <em>k</em> ² - <em>k</em> is even. This means that <em>k</em> ² - <em>k</em> = 2<em>m</em> for some integer <em>m</em>.
Induction step: show that the assumption implies (<em>k</em> + 1)² - (<em>k</em> + 1) is also even. We have
(<em>k</em> + 1)² - (<em>k</em> + 1) = <em>k</em> ² + 2<em>k</em> + 1 - <em>k</em> - 1
… = (<em>k</em> ² - <em>k</em>) + 2<em>k</em>
… = 2<em>m</em> + 2<em>k</em>
… = 2 (<em>m</em> + <em>k</em>)
which is clearly even. QED
To solve this find the common number that can be divided by both and use that number.
In this case, 4 is divisible by both 4 and 12, and thus the final expression would be
4(d + 3e).
Answer:
Twice a number decressed by five
