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.

I would appreciate if somebody could help me with the following problem

Q: Seeking a combinatorial proof

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

share|improve this question
1  
What do you consider to be a combinatorial proof? –  Gerry Myerson Oct 6 '13 at 2:07
    
Related - if not dup: math.stackexchange.com/questions/136237/… –  leonbloy Oct 6 '13 at 2:13
add comment

2 Answers

Consider a bag with balls numbered from $1$ to $n$. Number of ways of choosing $2$ balls with replacement is $n^2$.

We can also count the same in a different way. The pair of balls can be represented as $(i,j)$. Let us now look at the number of ways such that $\max\{i,j\} = k$, where $k \in \{1,2,\ldots,n\}$. If $C_k$ denotes the number of ways such that $\max\{i,j\} = k$, we then have $$C_k = 2k-1$$ This is because if $\max\{i,j\} = k$, then either $j<i=k$ or $i<j=k$ or $i=j=k$.

  • Number of ways such that $j<i=k$ is $k-1$.
  • Number of ways such that $i<j=k$ is $k-1$.
  • Number of ways such that $i=j=k$ is $1$.

Hence, we have $C_k=2k-1$. Hence, the total number of ways of choosing a pair of balls from $n$ balls with replacement is $$\sum_{k=1}^n C_k$$

share|improve this answer
4  
This is what I’d call a combinatorial proof. @Gerry’s familiar proof without words is elegant and non-computational, and I don’t have a really strong objection to calling it combinatorial, but I don’t think of it in those terms. (+1) –  Brian M. Scott Oct 6 '13 at 2:28
add comment

$$\matrix{a&b&c&d&e&\dots\cr b&b&c&d&e&\dots\cr c&c&c&d&e&\dots\cr d&d&d&d&e&\dots\cr e&e&e&e&e&\dots\cr\vdots&\vdots&\vdots&\vdots&\vdots&\dots\cr}$$

share|improve this answer
add comment

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.