the sum of consecutive odd numbers If the sum of consecutive odd numbers starting with $-3$ until $2k+1$ equals $21$
What is the value of $k$ ?
I can solve this by trying the numbers $-3-1+1+3+5+7+9=21$ , so the last term is $7th$ so the $k$ value is $3$
But I could not solve this with formula, I know the odd numbers come in the form of $2k+1$ but could not get much further.
 A: Notice that
$$\begin{align}1^2&=1\\2^2&=1+3\\3^2&=1+3+5\\\vdots&\qquad\qquad\qquad\quad\ddots\\k^2&=1+3+5+\dots+(2k-1)\\(k+1)^2&=1+3+5+\dots+(2k-1)+(2k+1)\end{align}$$
Thus,
$$-3-1+1+3+\dots+(2k+1)=(k+1)^2-4$$
A: If you are not aware of the result regarding sum of first odd numbers. An alternative is to note that this is an arithmetic progression.
This is an arithmetic series with the first term being $a=-3$ and the common difference $d=2$ and the last term being $2k+1$.
First let's figure out how many terms are there.
$$a+(n-1)d=2k+1$$
$$-3+2(n-1)=2k+1$$
$$2(n-1)=2k+4$$
$$n=k+3$$
Hence we have $\frac{n}{2}(a+2k+1)=21$
$$\frac{k+3}{2}(-3+2k+1)=21$$
$$\frac{k+3}{2}(2k-2)=21$$
$$(k+3)(k-1)=21$$
$$k^2+2k-24=0$$
$$(k-4)(k+6)=0$$
$k=4$ or $k=-6$ 
since the sequence increases, $k=4$.
A: The first $n$ natural number odds is $n^2$, so just manually add the $-3+-1$ and you get $n^2-4=21$ so $n^2=25$ or $n=5$.
A: The sum of the $N$ first non-zero natural numbers is equal to $\frac{N(N+1)}{2}$. It can be easily proven by induction: it is obviously true for $N=1$. 
Now, suppose it is for $N-1$, then 
$$\sum_{k=1}^{N}k=N+\sum_{k=1}^{N-1}k=N+\frac{(N-1)N}{2}$$ 
by induction hypothesis. Then, we deduce that 
$$\sum_{k=1}^{N}k=N(1+\frac{N-1}{2})=\frac{N(N+1)}{2}$$
as we claimed.
Now, let's remark that the sum of the $N$ first odd natural numbers is $$\sum_{k=1}^{N}(2k-1)=2\sum_{k=1}^{N}k-\sum_{k=1}^{N}1=2\frac{N(N+1)}{2}-N=N^{2}$$
Additionnally, $\sum_{k=1}^{N+1}(2k-1)=1+3+\dots+(2N+1)=(N+1)^{2}$
Thus, $$-3+(-1)+\underbrace{1+\dots+2N+1}_{=\sum_{k=1}^{N+1}(2k-1)}=-4+(N+1)^{2}=21$$ and we deduce that $N=4$.
A: You can reduce this sequence: $-3-1+1+3+5+7+9=21$.
$-3-1+1+3=0$. So you are being left with:
$5+7+9=21$. Since the last term is $9$ and the last term in your question is $2k+1$ we have:
$2k+1=9$
$2k=9-1$
$k=8:2$
$k=4$
Edit:
I would like to add, that your confusion might come from how you are understand "until $2k+1$". One might say: until some term or until some number. In the first instance you are right: $k=3$. In the second: $k=4$. Two following steps can help:
We can write an Arithmetic Sequence as a rule:
$X_n=a_1 + d(n-1)$
$a_1$ - first term
$d$ - common difference
$n$ - number of terms
Since you have $7$ terms, last term is equal:
$X_7=-3+(7-1)2=-3+12=9$
Before you can calculate the above first we need to find this $7$, therefore to sum up the terms of this arithmetic sequence you can use formula:
$\sum_{k=1}^{n}(a+kd)=\frac{n}{2}(2a+(n-1)d)$
$21=\frac{n}{2}(-6+2n-2)$
$21=-3n+n^2-n$
$n^2-4n-21=0$
Δ $=16-4*1*(-21)=16+84=100$
$n_1=7$ and $n_2=-3$
A: If you are in high school than you are familiar with arithmetic progressions which means a sequence is increasing with a fixed difference.
For example , $a,a+d,a+2d,a+3d\cdots,a+(n-2)d,a+(n-1)d$ is an A.P. of n terms 
Clearly $n^{th}$ term = $a+(n-1)d$
Your A.P. is $-3,-1,1,...,-3+(2k+1-1)2$
You have given the sum of A.P. which is 21.
The sum of n terms of a general A.P. is
$S_n=\frac{n(a+a_n)}{2}$
where $a_n $ means $n^{th}$ term
Putting values you will get 
$$21=\frac{(2k+1)(-3+(-3+(2k+1-1)2))}{2}$$$$42=(2k+1)(-6+4k)$$$$21=(2k+1)(2k-3)$$$$21=4k^2-4k-3$$$$4k^2-4k-24=0$$$$k^2-k-6=0$$$$(k-3)(k+2)=0$$Hence  $$k=3$$ or $$k=-2$$But k is no. of terms so it should be positive.
Hence you get $k=3$
$n^{th} $ term =$9$ 
A: Hint:
Have a look at the partial sums:
$$-3,-4,-3,0,5,12,21\cdots$$
Their plot follows a parabola.
If you add $4$ to get all positive values,
$$1,0,1,4,9,16,25\cdots$$ should ring a bell...
