3
$\begingroup$

The question asks to verify that each equation is true for every positive integer n.

The question is as follows:

$$1+ 3 + 5 + \cdots + (2n - 1) = n^2$$

I have solved the base step which is where $n = 1$.

However now once I proceed to the inductive step, I get a little lost on where to go next:

Assuming that k is true (k = n), solve for k+1:  (2k - 1) + (2(k+1) - 1) (2k - 1) + (2k+2 - 1) (2k - 1) + (2k + 1) 

This is where I am stuck. Do I factor these further to obtain a polynomial of some sort? Or am I missing something?

  • 0
    "Solve for" is definitely the wrong term. If I have the equation $x+y=5$ and I _solve for_ $y$, I get $y=5-x$. That's how that term should be used. If you write "Assuming that case $k$ is true, prove case $k+1$", then it would make sense.2012-03-08
  • 0
    I understand now. Next time I'll post the question with the correct terminology.2012-03-08

3 Answers 3

3

Assume true for $k$. Then consider the case $k+1$, you got $$1+3+\cdots+(2k-1)+(2(k+1)-1)$$ which is equal by inductive hypothesis $$k^2+(2k+1)=(k+1)^2$$ and this closes the induction.

  • 1
    As Michael Hardy said. You dont "solve". You do your base step, and then assume it holds for an arbitrary $k$, with this you have to show that $k+1$ holds as well.2012-03-08
  • 0
    So just to be clear, since I have proven that (2n - 1) is equivalent to n^2, I can just use n^2 + (2k + 1) and simplify to prove the next step in the induction case?2012-03-08
  • 0
    First of all $2n-1$ is not equivalent to $n^2$, the summation of odd numbers up to and including $2n-1$ equals to $n^2$, and the way induction works is: Prove it for a base case, then assume that it holds for an arbritrary number $n$, and then you have to use this to prove that the condition holds for $n+1$ as well. Think of it as dominos. Your base case is saying "the first domino falls" and then your inductive step is saying, suppose one domino falls, (show some work) then the next domino falls. Hence, all dominos fall.2012-03-08
  • 0
    Thanks. I need to be more careful with how I word things. I understood that the summation of odd numbers up to and including 2n-1 was n^2, I just phrased it wrong. Thanks. I think the domino effect actually makes it much clearer to me. My main problem was figuring out how to finish the problem. You have really helped me a lot though. thank you!2012-03-08
  • 0
    This blog helped me also: https://perkss.github.io/#/MathFundamentals/MathPreliminaries#text-body2018-01-13
2

You have $$ 1+3+5+7+\cdots+(2k-1)\quad + \quad \Big( 2(k+1) - 1 \Big). $$ This is equal to $$ k^2 \quad + \quad \Big( 2(k+1) - 1\Big). $$ Simplify: $$ k^2 + 2k + 2 - 1 $$ $$ = k^2 + 2k + 1 $$ $$ = (k+1)^2. $$

2

Even though there is an accepted answer, I feel compelled to give a "geometric proof," and by that, I mean a picture. This is in the spirit of discrete mathematics' counting in two ways. 4 squares, each one larger than the previous by one unit side length, with the addition in blue.

The small blue squares of any individual larger square represent the last term of your sum. Here, I have shown $n=1,2,3,4$. Obviously, each large square has $n^2$ area. Or you can see that it can also be thought of as adding up the "L" shaped additions. Since both of these methods are counting the same thing, they must be the same number.