# Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$

Prove that for all integers $n$, $n \geq 1$, $$1 + 3 + 5 + \cdots + (2n - 1) = n\cdot n$$

How would I go about proving this?

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Normally, just use induction. Alternatively, show that $1 + (2n-1) + 3 + (2n-3) + 5 + (2n-5) + \cdots + (2n-3) + 3 + (2n-1) + 1 = 2n \cdot n$. – Johannes Kloos Apr 24 '12 at 13:12
@J.M. I don't think Gauss invented that trick, there's just a famous story (possibly apocryphal) about Gauss as a child finding that method. – Thomas Andrews Apr 24 '12 at 13:18
Hint $\$ By telescopy it's equivalent to $\rm (n+1)^2 - n^2 = 2\:n+1.\:$ See Eric's answer for a geometric view. – Gone Apr 24 '12 at 13:52
Sometimes unmotivated questions without any work shown by the OP get all the love... – The Chaz 2.0 Apr 24 '12 at 20:57
cuz its easy and people get to name-drop Gauss? who knows. – Michael Edenfield Apr 24 '12 at 22:55

(1) Proof using the "method of gauss":

$$\begin{eqnarray*} 2(1 + 3+ 5 + \ldots (2n-1)) &=&\big[ 1 + (2n-1)\big] + \big[3 + (2n-3)\big] + \ldots + \big[ (2n-1) + 1\big] \\ &=& \underbrace{2n + 2n + \ldots + 2n}_{\text{n times}} \\ &= &2n(n). \end{eqnarray*}$$ Therefore it follows that $(1 + 3 + \ldots (2n-1)) = n\times n.$

(2) Proof by induction: Let $P(n)$ be the statement:

"For all positive integers $n$, $1 + 3 + \ldots + (2n-1) = n^2$."

For $n=1$ clearly $(2 \cdot 1) - 1 = 1\cdot 1$ so that $P(1)$ is true. So suppose the result holds for $n=k$, i.e. $P(k)$ is true. Since $P(k)$ is true, this means that we have the following equality when $n=k$:

$$1+ 3 + \ldots + (2k-1) = k^2.$$

If you add $2k+1$ to both sides of this equation, you get that

$$\begin{eqnarray*} 1+ 3 + \ldots +(2k-1) + (2k+1) &=& k\cdot k + (2k+1) \\ &=& k^2 + 2k + 1 \\ &=& (k+1)(k+1) \end{eqnarray*}$$

showing that the result holds for $n=k+1$, i.e. $P(k+1)$ is true. Therefore by the Principle of Mathematical Induction, $1 + 3 + \ldots + (2n-1) = n\cdot n$ for all positive integers $n$.

(3) Suppose you only knew that the sum of $n$ consecutive integers is $n(n+1)/2$. Then

$$\begin{eqnarray*}1 +2 + \ldots + 2n &=& \frac{2n(2n+1)}{2}\\ \implies 1 + 3 + \ldots + (2n-1) &=& n(2n+1) - 2(1 + \ldots + n) \\ &=& 2n^2 + n - 2\left(\frac{n(n+1)}{2}\right) \\ &=& 2n^2 + n - n^2 - n \\ &=& n^2. \end{eqnarray*}$$

(4) Proof using telescoping sums (idea by Bill Dubuque):

$$\begin{eqnarray*} \sum_{k=1}^n 2k-1 &=& \sum_{k=1}^n k^2 - (k-1)^2 \\ &=& (1 - 0) + (2^2 -1 ) + \ldots + n^2 - (n-1)^ 2\\ &=& 1 + (-1 + 2^2) + (-2^2 + 3^2) + \ldots + (-(n-2)^2 + (n-1)^2) + (-(n-1)^2 + n^2)) \\ &=& n^2. \end{eqnarray*}$$

(5) Proof by method of differences (brute force): Let $a_n = \sum_{k=1}^n 2k-1$. Then we see that $a_1 =1$, $a_2 = 4$, $a_3 = 9$, etc.

Now we look at the first differences $4 - 1 = 3$, $9- 4 = 5$, $16 - 9 = 7$, etc. Then when we look at the second difference, notice that it is constant: $5-3 = 2$, $7- 5= 2$, etc so we conjecture that

$$\sum_{k=1}^n 2k - 1 = an^2 + bn + c$$

where $a,b,c$ are constants to be determined. Plugging in $n = 1,2,3$ gives us a $3 \times 3$ linear system to solve, namely the linear system

$$\left[\begin{array}{ccc} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 9 & 3 & 1 \end{array}\right]\left[\begin{array}{c} a \\ b \\ c \end{array}\right] = \left[\begin{array}{c} 1 \\ 4 \\ 9 \end{array}\right].$$

The determinant of the coefficient matrix is

$$\begin{eqnarray*} 1(2-3) - 1(4-9) + 1(12 - 81) &=& -1 + 5 - 69 \\ &\neq& 0 \end{eqnarray*}$$ so we have a unique solution. It is easy to see that $a = 1, b= 0, c=0$ is a solution to the linear system above. By the previous line, it is the only solution so we are done.

(6) Proof by a direct bash: Suppose you only knew that the sum of the first $n$ integers is $n(n+1)/2$. Then

$$\begin{eqnarray*} \sum_{k=1}^n 2k - 1 &=& \bigg(2\sum_{k=1}^n k \bigg) - \sum_{k=1}^n 1 \\ &=& 2\left(\frac{n(n+1)}{2}\right) - n\\ &=& n^2 + n -n\\ &=& n^2 \end{eqnarray*}.$$

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Could someone say more about the construction of the coefficient matrix in (5)? – Tony May 7 '12 at 13:50
@Tony You plug in $n=1,2,3$ into the formula I gave above and you get the equations $a + b + c = 1$, $4a + 2b + c = 4$, and $9a +3b + c = 9$. I just put these simultaneous equations into matrix form. – BenjaLim May 7 '12 at 23:02
+1 for proof by direct bash – Chris Dugale Jul 26 '12 at 6:39

Hint: When $n=5$, $$\boldsymbol{1+{\color{red}3}+{\color{green}5}+{\color{purple}7}+{\color{blue}9}=}$$ $$\begin{array}{ccccc} \blacksquare & {\color{red}\blacksquare} & {\color{green}\blacksquare} & {\color{purple}\blacksquare} & {\color{blue}\blacksquare}\\ {\color{red}\blacksquare} & {\color{red}\blacksquare} & {\color{green}\blacksquare} & {\color{purple}\blacksquare} & {\color{blue}\blacksquare}\\ {\color{green}\blacksquare} & {\color{green}\blacksquare} & {\color{green}\blacksquare} & {\color{purple}\blacksquare} & {\color{blue}\blacksquare}\\ {\color{purple}\blacksquare} & {\color{purple}\blacksquare} & {\color{purple}\blacksquare} & {\color{purple}\blacksquare} & {\color{blue}\blacksquare}\\ {\color{blue}\blacksquare} & {\color{blue}\blacksquare} & {\color{blue}\blacksquare} & {\color{blue}\blacksquare} & {\color{blue}\blacksquare} \end{array}$$

$$\boldsymbol{=5^2}.$$

Turn this into a general proof.

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I'm pretty sure the associated diagram has already been posted somewhere in this site... – J. M. Apr 24 '12 at 13:21
@J.M. I made my own in TeX with matrices and colors. How does it look? (I don't know where to find the original image) – Eric Apr 24 '12 at 13:26
One doesn't have to click edit, just right click on the formula and choose "Show math as TeX commands". – Hans Lundmark Apr 24 '12 at 14:31
By the way, interesting idea to use the "blacksquare" command for all those colored boxes. ;-) – Hans Lundmark Apr 24 '12 at 14:33
I am going to teach this to my young school students ! :) – AJed Jan 11 at 4:32

The method of Gauss, graphically:

BTW: this corresponds to proof 1 in Benjamin's answer; and the graphic of Eric's answer corresponds to proof 4 (telescoping squares).

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$$\begin{eqnarray*} (1 + 3 + 5 + ... + (2n - 1)) &=& (1 + 2 + 3 + ... + 2n) - (2 + 4 + ... + 2n)\\ &=& \sum_{i=1}^{2n}{i} - 2\sum_{i=1}^{n}{i}\\ &=& \frac{(2n)(2n+1)}{2} - 2\left(\frac{n(n+1)}{2}\right)\\ &=& n(2n+1) - n(n+1)\\ &=& n(2n + 1 - n - 1) = n^{2}\end{eqnarray*}$$

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By inspection note that $1 = 1 \cdot 1$, that $1 + 3 = 2 \cdot 2$, and that $1 + 3 + 5 = 3 \cdot 3$.

Next, witness that the difference between any two successive perfect squares $(n - 1)^2$ and $n^2$ is a particular odd integer:

$n^2 - (n - 1)^2 \implies n^2 - (n^2 - 2n + 1) \implies 2n - 1$

Note that this is the very formula which you are summing to generate your series.

So for instance the difference between $3 \cdot 3$ and $2 \cdot 2$ is $2 \cdot 3 - 1 = 5$, corresponding to the $5$ in $3\cdot3 = 1 + 3 + 5$.

If the sum of the series generated so far is a square (shown by inspection), and the next term is from the series generated by $2n - 1$, then after this term is added, the sum must be the next square because the each term obeys the formula for the difference between two successive squares.

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Note that $(n+1)^2=n^2+2n+1$. This defines a recursive relationship between $n+1$ and $n$. This means that we can write the function $f(x)=x^2$ recursively as: $f(x)=f(x-1)+2x-1$ with $f(1)=1$. Now, simply expand out $f(x)$ to obtain $f(n)=\sum_{k=1}^{n}2x-1=n^2$.

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Consider this as the sum of a finite recursive series. Each element in the series is calculated from a fixed point a1=1 by adding 2 to the previous element. So, the equation for any arbitrary an is

$$a_{n} = [a_{n-1} + 2, a_{1} = 1] = 1 + 2(n-1)$$

The sum of n terms of this recursive series breaks down to:

$$\sum_{x=1}^{n}[ a_{x-1}+2, a_{1}=1] = 1 + (1 + 2) + ((1 + 2) + 2) + (((1 + 2) + 2) + 2) + \cdot\cdot\cdot + (1 + 2(n-1))$$

By inspection, we see that 1 occurs n times, once per term, and 2 occurs in a triangular fashion (each term contains one more 2 than the previous term). So, this condenses to the quantity of n plus double the (n-1)th triangular number (as the first term has zero 2s):

$$\sum_{x=1}^{n}( a_{x-1}+2, a_{1}=1) = n + 2\frac{n(n-1)}{2}$$ $$= n(n-1) + n$$ $$= n^2 - n + n$$ $$= n^2$$

QED.

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you can see this link in generally solution :

http://en.wikipedia.org/wiki/Arithmetic_progression

This section is about Finite arithmetic series. For Infinite arithmetic series, see Infinite arithmetic series.

The sum of the members of a finite arithmetic progression is called an arithmetic series.

Expressing the arithmetic series in two different ways:

$S_n=a_1+(a_1+d)+(a_1+2d)+\cdots+(a_1+(n-2)d)+(a_1+(n-1)d)\\ S_n=(a_n-(n-1)d)+(a_n-(n-2)d)+\cdots+(a_n-2d)+(a_n-d)+a_n.$

Adding both sides of the two equations, all terms involving d cancel:

$\ 2S_n=n(a_1+a_n).$

Dividing both sides by 2 produces a common form of the equation:

$S_n=\frac{n}{2}( a_1 + a_n).$

An alternate form results from re-inserting the substitution:$a_n = a_1 + (n-1)d:$

$S_n=\frac{n}{2}(2a_1 + (n-1)d).$

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