# Proving $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$

Prove $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$.

So the provided solution avoids induction and makes use of the fact that $1 + 3 + 5 + \cdots + (2n-1) = n^{2}$ however I cannot understand the first step: $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1) = (1 + 3 + 5 + \cdots + (4n-1)) -(1 + 3 + 5 + \cdots + (2n-1))$. Once that has been established I can follow the rest, but I was hoping someone could help me understand why $(1 + 3 + 5 + \cdots + (4n-1))$ which already looks like less than the LHS can be made equal to the LHS by subtracting a positive number.

Additionally, I wanted to prove the equality using induction, but had trouble with that as well. I think I am thrown off by the last term $4n-1$. Even the initial case of $n=1$ is not totally clear to me: does it hold because $2(1)+1 = 3(1)^{2}$ or is it because $4(1)-1 = 3(1)^{2}$?

Either way, my approach was to replace every $n$ on the LHS with $n+1$ which resulted in: $$(2(n+1)+1) + (2(n+1)+3)+ \cdots + (4(n+1)-1)$$ and I am not sure what the second to last term in the sequence would be... I tried simplifying anyway $$(2n+3) + (2n+5) + \cdots + (4n+3)$$ and I this point I thought I could subtract $(2n+1)$ from both sides of the induction assumption resulting in $(2n+3) + (2n+5) + \cdots ? = 3n^{2} - (2n+1) - (4n-1)$ substituting this yields: $$3n^{2} - 2n - 1 - 4n + 1 + 4n + 3 = 3n^{2} + 6n + 3 - 8n + 1 = 3(n+1)^{2} - 8n + 1$$ but I guess I don't want the $-8n + 1$... I also tried substituting $3n^{2} - (2n+1)$ without that last term being subtracted, but that did not work out either. If anyone can help me understand how to do this properly I would really appreciate it. Thanks!

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The first item, $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1)$

$= (1 + 3 + 5 + \cdots + (4n-1)) -(1 + 3 + 5 + \cdots + (2n-1))$

comes because you are just adding and subtracting the same set of terms on the RHS.

Your base case for the induction has just one term on the left side, which I would write $2*1+1=3*1^2$ For the induction, assume $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1)=3n^2$ then extend it to $n+1$: $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1) + (4n+1)+(4n+3)-(2n+1)$

$=3n^2+4n+1+4n+3-(2n+1)=3n^2+6n+1=3(n+1)^2$

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Unfortunately I still don't get it... with the induction, is it correct to sort of 'plug in' $n+1$ for each $n$? And somewhere in between $(2n+5)$ and $(4n-1)$ each term begins to be calculated differently? I can see why $2n+1$ is some odd number, but what is $4n-1$? –  ghshtalt Mar 7 '11 at 22:31
No, with the induction you figure out which terms are added (and in this case, which is not so normal, deleted) when you go from n to n+1. So the bottom term when using n is 2n+1. When using n+1 it is 2(n+1)+1, so you lose 2n+1. The upper limit goes from 4n-1 to 4(n+1)-1=4n+3, so you gain the two terms I show. –  Ross Millikan Mar 7 '11 at 22:50
If you think of it as the sum by 2's from (2n+1) through (4n-1), then when you consider n+1 you go from 2(n+1)+1=2n+3 to 4(n+1)-1=4n+3 –  Ross Millikan Mar 7 '11 at 22:52
thank you for the reply, I can understand much more now. Just to be absolutely clear: in order to determine which terms are added/deleted one could replace each $n$ in the original LHS with $n+1$, correct? –  ghshtalt Mar 7 '11 at 22:57
No, because there is one more term in the n+1 case than in the n case. For example, if n=9, you add the odd numbers from 19 through 35 (9 terms). When n=10, you add the odd numbers from 21 through 39 (10 terms). I don't see how to "change 9 to 10 in each term" –  Ross Millikan Mar 7 '11 at 23:06

$\rm\ S(n) := \sum_{k=0}^{n-1}\ 2k+1 = 1 + 3 +\:\cdots\:+ 2n-1\ = n^2\$ so that it's clear that your sum is simply $\rm\ S(2\:n) - S(n) = 4n^2 - n^2 = 3n^2\:.\:$ Therefore you need only prove that the first sum holds true. Hint: the inductive step is equivalent to the fact that $\rm\ S(n+1) - S(n) = (n+1)^2 - n^2 = 2n+1\:.\:$

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Thank you for the answer, although I must say that it isn't the same without the bold HINT formatting ;) –  ghshtalt Mar 7 '11 at 23:13

Your initial problem seems to be seeing $$1 + 3 + 5 + \cdots + (4n-1) = 1 + 3 + 5 + \cdots + (2n-1) + (2n+1) + (2n+3) + \cdots + (4n-1)$$ so the LHS is $(2n)^2 = 4n^2$ and the left half of the RHS is $n^2$, leaving $3n^2$ for the right half of the RHS.

For induction, you need to start with one term where $(2 \times 1 +1) = 3 \times 1^2$, and then given $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^2$ that you can prove $(2n+3) + \cdots + (4n-1) + (4n+1) + (4n+3) = 3(n+1)^2$, i.e. that $3n^2 - (2n+1) + (4n+1) + (4n+3) = 3(n+1)^2$, which is not difficult.

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Definitely helps a lot, thank you! –  ghshtalt Mar 7 '11 at 23:10

Gauss gives this solution:

$$(4n-1+2n+1)*\frac{(\frac{(4n-1-(2n+1))}{2}+1)}{2}$$

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The simple solution using $1+3+\ldots+(2n-1)=n^2$ is as follows:

$(2n+1)+(2n+3)+\ldots+(4n-1)=2n\cdot n+ (1+3+\ldots+(2n-1))=3n^2$

The fact that $1+3+\ldots+(2n-1)=n^2$ is true is a simple induction: Assume it's true for $n$. Then for $n+1$ we get $1+3+\ldots+(2n-1)+(2(n+1)-1)=n^2+2n+1=(n+1)^2$

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Overall:

\begin{array}{cccccc} \color{red}{\bullet}&\color{orange}{\bullet}&\color{yellow}{\bullet}&\color{green}{\bullet}&\color{blue}{\bullet}&\color{purple}{\bullet}&\\ \color{orange}{\bullet}&\color{orange}{\bullet}&\color{yellow}{\bullet}&\color{green}{\bullet}&\color{blue}{\bullet}&\color{purple}{\bullet}&\\ \color{yellow}{\bullet}&\color{yellow}{\bullet}&\color{yellow}{\bullet}&\color{green}{\bullet}&\color{blue}{\bullet}&\color{purple}{\bullet}&\\ \color{green}{\bullet}&\color{green}{\bullet}&\color{green}{\bullet}&\color{green}{\bullet}&\color{blue}{\bullet}&\color{purple}{\bullet}&\\ \color{blue}{\bullet}&\color{blue}{\bullet}&\color{blue}{\bullet}&\color{blue}{\bullet}&\color{blue}{\bullet}&\color{purple}{\bullet}\\ \color{purple}{\bullet}&\color{purple}{\bullet}&\color{purple}{\bullet}&\color{purple}{\bullet}&\color{purple}{\bullet}&\color{purple}{\bullet}\\ \end{array}

$$\color{red}{1}+\color{orange}{2}+\cdots+\color{blue}{4n-3}+\color{purple}{4n-1}=(2n)^2$$

\begin{array}{ccc|ccc} &&&\color{green}{\bullet}&\color{blue}{\bullet}&\color{purple}{\bullet}&\\ &&&\color{green}{\bullet}&\color{blue}{\bullet}&\color{purple}{\bullet}&\\ &&&\color{green}{\bullet}&\color{blue}{\bullet}&\color{purple}{\bullet}&\\ \hline \color{green}{\bullet}&\color{green}{\bullet}&\color{green}{\bullet}&\color{green}{\bullet}&\color{blue}{\bullet}&\color{purple}{\bullet}&\\ \color{blue}{\bullet}&\color{blue}{\bullet}&\color{blue}{\bullet}&\color{blue}{\bullet}&\color{blue}{\bullet}&\color{purple}{\bullet}\\ \color{purple}{\bullet}&\color{purple}{\bullet}&\color{purple}{\bullet}&\color{purple}{\bullet}&\color{purple}{\bullet}&\color{purple}{\bullet}\\ \end{array}

$$\color{green}{2n+1}+\cdots+\color{blue}{4n-3}+\color{purple}{4n-1}=3n^2$$

But to your first question, $1+3+\cdots+(4n-1)$ is definitely larger than $(2n+1)+(2n+3)+\cdots+(4n-1)$, because the second sum is the same as the first, except that it is missing the first few terms.

\begin{align} 1+3+\cdots+(4n-1)&=\Big(1+3+\cdots+(2n-1)\Big)+\Big((2n+1)+(2n+3)+\cdots+(4n-1)\Big) \end{align}

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