Mathematical Induction
G’day im currently working on a Math C assignment it consists of determining a formula for a rectangular Patten of tiles, then proving it by mathematical induction the formula is:
2n²-2n+1
However no mater how hard I try I can’t work out how to prove it. Any help would be much appreciated
thanks Quill
thanks Dan These are the Questions
Question 1
By considering the tile patterns shown and other patterns in this design, determine a formula for the number of tiles in the nth pattern.
(the colour has nothing to do with the patten just shoes each square)
Question 2
Prove the formula you obtained for the nth pattern.
The answer i got for Question one is 2n²-2n+1 i got it by dividing the patterns up in to equal proportions
then i worked out each section like a rectangle (n x (n-1))/2
multiplied that by 4 to account for each section then added 1 for the center one. it ended up like this:
then with a little rearrangement it became((n x (n-1))/2) x 4 + 1
its just Question 2 i don't understand how to do.2n²-2n+1
Here is a solution using a recursive function.
Define and for we get , where is the number of rectangles in each pattern.
To see how this works, consider what we may call the “main line”. The number of rectangles this “main line” is . We a to add 2 rectangles, one above and one below, each of the rectangles on the main line plus one on each end.
Thus : we start with the number we have add two for each on the main line plus two more, one on each end.
It can be shown that the closed form is .
Using the formula , look at picture 3:
Starting from picture 2, we add 2 rectangles above each square along the main line (including the middle one which already has two squares attached to it). Then we add 2 rectangles to the end to get the third picture.
Thanks Plato and tukeywilliams I think I got it does this sound right
Proving Via Mathematical Induction
2k² - 2k + 1
Explain how that Works Sn+1 = Sn + 2(2n – 1) +2
Then prove that 2k² - 2k + 1 Works when n = 1
2 x 1² - 2 x 1 + 1 = 1
Then Prove that it works when n = k + 1
2(k+1)² - 2(k+1) + 1 = 2k² - 2k + 1 + 4k2(k² + 2k + 2) –2k + 2 +1 = 2k² + 2k +12k² + 4k + 4 –2k + 3 = 2k² + 2k +12k² + 2k + 1 =2k² + 2k +1
Proof: We use induction on . Base case: For , . Inductive step: Suppose now as inductive hypothesis that is true for some positive integer . Then as required. Conclusion: Hence, by induction, is true for all positive integers .
Your second line followed incorrectly from your first line (i.e. ).