Science:Math Exam Resources/Courses/MATH220/December 2011/Question 06 (a)

MATH220 December 2011

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q2 (a)  •  Q2 (b)  •  Q3  •  Q4  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q8  •  

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Question 06 (a)
Prove that $1+3+5+\ldots +(2n-1)$ is a perfect square for all n ∈ N. (An integer m is a perfect square if m = k2 for some integer k.)

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
You should try to conjecture a formula for the sum 1 + 3 + 5 +... + (2n-1) and prove that formula holds by induction.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We start our induction at n = 1. In that case we have that 2n - 1 = 1 and so the sum is just 1 alone which is a perfect square. We can check n = 2 if we want as well, in that case 2n - 1 = 4 - 1 = 3 and so the sum is 1 + 3 = 4 which is the square of 2. For n = 3 we have that 1 + 3 + 5 = 9 which is the square of 3; we now see something which seems to be a pattern, that is for n = k the sum will end up being the square of the number k. And so we conjecture that $\displaystyle 1+3+5+\ldots +(2n-1)=n^{2}$ We have proved above that this is true for n = 1, 2 and 3. We will use induction to prove that is holds for all values of n. Since we have the first step(s) done, we now assume that the conjecture holds for n = k and we will show that it still holds for n = k+1. Indeed ${\begin{aligned}\underbrace {1+3+5+\ldots +(2k-1)} _{=k^{2}}+(2(k+1)-1)&=k^{2}+(2k+2-1)\\&=k^{2}+2k+1\\&=(k+1)^{2}\end{aligned}}$ This shows that our conjecture is true for n = k+1 and thus concludes our proof.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The following proof is not valid for the purpose of this exam question since it doesn't use induction, but it is of the level of this course, so we add it here for completeness. Consider playing with square pieces of paper, or colouring squares in a lined piece of paper. If you look at a n by n square, it is made of n2 squares. The next square you can built, which will have now n+1 squares on each side will contain (n+1)2 squares, but how many squares were added to it? Well you need to add a layer on say the right side, so n additional squares and also a layer on the top side, so another n additional squares and now you are also missing one square at the top right corner, so yet another square. In total, to go from the square with n little squares on the side to the next one with n+1 squares on each side you need to add 2n+1 squares. So if we start with 1 square, we need to add 3 squares to get a 2 by 2 squares and we need to add 5 squares to that to get a 3 by 3 square and we need to add 7 squares to get a 4 by 4 square and hence if we sum $1+3+5+\ldots +(2n-1)$ squares we will get a n by n square. Which is exactly what we wanted to show.

Solution 3

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The following proof is not valid for the purpose of this exam question since it doesn't use induction, but it is of the level of this course, so we add it here for completeness. Another proof can be obtained by simply playing with the formula that sums consecutive integers, that is $1+2+3+4+5+6+\dots +k=\sum _{i=1}^{k}i={\frac {k(k+1)}{2}}$ And we just use the fact that ${\begin{aligned}1+3+5+\dots +(2n-1)&=\left(1+2+3+\ldots +(2n-1)+(2n)\right)\\&\quad -(2+4+6+\ldots +(2n))\end{aligned}}$ Or if we use sum notations $\sum _{i=1}^{n}(2i-1)=\sum _{i=1}^{2n}i-\sum _{i=1}^{n}2i$ Now using the formula above we obtain that $\sum _{i=1}^{2n}i={\frac {2n(2n+1)}{2}}=n(2n+1)=2n^{2}+n$ and $\sum _{i=1}^{n}2i=2\sum _{i=1}^{n}i=2{\frac {n(n+1)}{2}}=n(n+1)=n^{2}+n$ And so $\displaystyle 1+3+5+\dots +(2n-1)=(2n^{2}+n)-(n^{2}+n)=n^{2}$ which proves that it is a perfect square.

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