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
Answer:
132
Step-by-step explanation:
the top rectangle is 28
the middle rectangle is 56
each of the triangles are 24
48+28+56
Answer:
-21 is the answer , hope it helps
Step-by-step explanation:
Answer:
A and E
Step-by-step explanation:
Given
Graphs A to E
Required
Which do not have solutions
When there are no point of intersection between lines and/or curves of a graph, then such graph has no solution.
Using the above description as a yard stick, the first (A) and the last (E) graph have no solution.
