Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
"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. –  Michael Hardy Mar 8 '12 at 4:39
    
I understand now. Next time I'll post the question with the correct terminology. –  TMGunter Mar 8 '12 at 4:42

3 Answers 3

up vote 3 down vote accepted

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.

share|improve this answer
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. –  Daniel Montealegre Mar 8 '12 at 4:42
    
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? –  TMGunter Mar 8 '12 at 4:44
    
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. –  Daniel Montealegre Mar 8 '12 at 4:48
    
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! –  TMGunter Mar 8 '12 at 4:52

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: $$ k^2 + 2k + 2 - 1 $$ $$ = k^2 + 2k + 1 $$ $$ = (k+1)^2. $$

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.