How to derive the formula for the sum of the first $n$ odd numbers: $n^2=\sum_{k=1}^n(2k-1).$ [duplicate]

Question

How to derive this formula?

$$n^2=\sum_{k=1}^n(2k-1).$$

Answer 1

Notice how the difference of two consecutive squares is always an odd number: $$(x+1)^2=x^2+(2x+1)$$

Answer 2

$$\sum_{k=1}^n(2k-1) = \sum_{k=1}^n 2k + \sum_{k=1}^n - 1 = 2\sum_{k=1}^nk - n = 2\dfrac{n(n+1)}{2} - n = n^2 $$

Answer 3

Well, since other had given formulae for calculating this sum, I will write a summing method from my primary school days: For $n=5$,

$$\begin{array}{c|c|c|c|c} \times&\times&\times&\times&\times\\\hline \color{red}\bigcirc&\color{red}\bigcirc&\color{red}\bigcirc&\color{red}\bigcirc&\times\\\hline \times&\times&\times&\color{red}\bigcirc&\times\\\hline \color{red}\bigcirc&\color{red}\bigcirc&\times&\color{red}\bigcirc&\times\\\hline \times&\color{red}\bigcirc&\times&\color{red}\bigcirc&\times \end{array}$$

Answer 4

Hint:

$$2(1)-1+2(2)-1+2(3)-1+\dots 2(n)-1$$ $$=2(1)+2(2)+2(3)+\dots+2(n)-(1+1+\dots 1)$$

$$=2(1+2+3+\dots+n)-n$$

At least now you should be able to do If you are familiar with the formula for :

sum of first $n$ natural numbers = ??

Answer 5

That depends on what you already know. One way to reduce this to a better known equality would be to use $1+3+5+7=(1+2+3+4+5+6+7)-2(1+2+3)$.

Answer 6

As stated above, it can be proven by Mathematical Induction or by simple algebraic manipulation. But there is other technique that is less rigorous but makes it obvious why it is true: 'Proof' by picture, look at the slide named "Sum of the Odd Integers": http://math.berkeley.edu/~rbayer/09su-55/handouts/ProofByPicture-printable.pdf

Answer 7

You could use mathematical induction to do this: First prove the statement for n=1 (or 0 if you like) and then prove that if it holds for n=k, it is also true for n=k+1. (You can do the last part by looking at differences between consecutive squares for example.)

Answer 8

Checking the first few cases, we find:

$1 = 1$

$1 + 3 = 4$

$1 + 3 + 5 = 9$

$1 + 3 + 5 + 7 = 16$

Okay: The sequence $1, 4, 9, 16, \ldots$ might be enough evidence to hazard a guess, namely, that the sum of the first $n$ odds is $n^2$.

Let's state this more formally in terms of a proposition. Given $n \in \mathbb{N}$, we write the following:

$P(n):$ the sum of the first $n$ odds is $n^2$.

This is a proposition about the natural numbers; so we can now proceed to see if it can be proved by mathematical induction. First up is the base case, $P(1)$. What does $P(1)$ mean in plain English?

$P(1):$ the sum of the first $1$ odd is $1^2$.

This is true: $1 = 1^2$.

Excellent; the base case is taken care of!

Next up is the inductive hypothesis (IH):

Suppose that $P(k)$ holds for $k \in \mathbb{N}$. Written out, this says that we assume the following:

$P(k):$ the sum of the first $k$ odds is $k^2$.

That is, $1 + 3 + \cdots + (2k-1) = k^2$.

To finish off the proof by induction we need to prove the following:

$P(k) \implies P(k+1)$

Stated in plain English, we need to show that if $P(k)$ is true, then $P(k+1)$ is true, too.

What does $P(k+1)$ assert? It says that:

$P(k+1):$ the sum of the first $k+1$ odds is $(k+1)^2$.

Is this true?

Well, the first $k+1$ odds are: $1 + 3 + \cdots + (2k-1) + (2k+1)$.

Note that, by our IH, we know the sum of the first $k$ of these addends; namely, we know that:

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

And so, by hypothesis, the sum of the first $k+1$ odds can be written as:

$1 + 3 + \cdots + (2k-1) + (2k+1) = k^2 + (2k+1) = k^2 + 2k + 1 = (k+1)^2$, where the final equality was just a matter of factoring the expression at hand.

A ha! In addition to $P(1)$ being true, we have now also proved that $P(k) \implies P(k+1)$.

And so, by the Principle of Mathematical Induction, $P(n)$ holds for all $n \in \mathbb{N}$. QED.

Answer 9

for k=1,2,3,3............n 2(1)−1+2(2)−1+2(3)−1+…2(n)−1 =2(1)+2(2)+2(3)+⋯+2(n)−(1+1+…1) =2(1+2+3+⋯+n)−n sum of first natural natural numbers:-1+2+3+........+n now if you can compute sum of first natural numbers, I don't think there would be any problem

Answer 10

First we define of the sum the integers from $1$ to $n$: $$ S(n) := \sum_{k=1}^n k $$ then the sum of the odd numbers from $1$ to $n$ (smaller or equal to $n$): $$ S_{\tiny\mbox{odd}}(n) := \sum_{k=1 \atop k {\tiny \mbox{odd}}}^n k = \sum_{k=1}^{\lceil n/2 \rceil} 2k+1 $$ and the sum of the even numbers from $1$ to $n$ (smaller or equal to $n$): $$ S_{\tiny\mbox{even}}(n) := \sum_{k=1 \atop k {\tiny \mbox{even}}}^n k = \sum_{k=1}^{\lfloor n/2 \rfloor} 2k $$ We assume $$ S(n) = S_{\tiny\mbox{odd}}(n) + S_{\tiny\mbox{even}}(n) \quad (*) $$ Further we notice $$ \begin{align} S(n+1) &= S(n) + n + 1 \\ S(n+2) &= S(n) + n + 1 + n + 2 \\ S(n+3) &= S(n) + n + 1 + n + 2 + n +3 \\ &\vdots \\ S(n+n) &= S(n) + n^2 + S(n) \end{align} $$ or $$ S(2n) = 2 S(n) + n^2 $$ On the other hand we notice: $$ 2 \, S(n) = S_{\tiny\mbox{even}}(2n) $$ so we get $$ S(2n) = n^2 + S_{\tiny\mbox{even}}(2n) $$ which with $(*)$ implies $$ S_{\tiny\mbox{odd}}(2n) = n^2 $$ with $$ S_{\tiny\mbox{odd}}(2n) = \sum_{k=1}^{\lceil 2n/2 \rceil} 2k+1 = \sum_{k=1}^n 2k+1 $$