Using a speed-time graph

The graph in the figure is a plot of speed versus time. Distance covered is the area under the line. Each time interval is coloured differently. The distance covered in the second and subsequent intervals is the area of its trapezium, which can be subdivided into triangles as shown. As each triangle has the same base and height, they have the same area as the triangle in the first interval. It can be observed that every interval has two more triangles than the previous one. Since the first interval has one triangle, this leads to the odd numbers.^[1]

Using the sum of first n odd numbers

From the equation for uniform linear acceleration, the distance covered

$${\displaystyle s=ut+{\tfrac {1}{2}}at^{2}}$$

for initial speed ${\displaystyle u=0,}$ constant acceleration ${\displaystyle a}$ (acceleration due to gravity without air resistance), and time elapsed ${\displaystyle t,}$ it follows that the distance ${\displaystyle s}$ is proportional to ${\displaystyle t^{2}}$ (in symbols, ${\displaystyle s\propto t^{2}}$), thus the distance from the starting point are consecutive squares for integer values of time elapsed. The middle figure in the diagram is a visual proof that the sum of the first ${\displaystyle n}$ odd numbers is ${\displaystyle n^{2}.}$^[2] In equations:

1	= 1	= 12
1 + 3	= 4	= 22
1 + 3 + 5	= 9	= 32
1 + 3 + 5 + 7	= 16	= 42
1 + 3 + 5 + 7 + 9	= 25	= 52

That the pattern continues forever can also be proven algebraically:

$${\displaystyle {\begin{aligned}\sum _{k=1}^{n}(2\,k-1)&={\frac {1}{2}}\,\left(\sum _{k=1}^{n}(2\,k-1)+\sum _{k=1}^{n}(2\,(n-k+1)-1)\right)\\&={\frac {1}{2}}\,\sum _{k=1}^{n}(2\,(n+1)-1-1)\\&=n^{2}\end{aligned}}}$$

To clarify this proof, since the ${\displaystyle n}$^th odd positive integer is ${\displaystyle m\,\colon =\,2n-1,}$ if ${\displaystyle S\,\colon =\,\sum _{k=1}^{n}(2\,k-1)\,=\,1+3+\cdots +(m-2)+m}$ denotes the sum of the first ${\displaystyle n}$ odd integers then

$${\displaystyle {\begin{alignedat}{4}S+S&=\;\;1&&+\;\;3&&\;+\cdots +(m-2)&&+\;\;m\\&+\;\;m&&+(m-2)&&\;+\cdots +\;\;3&&+\;\;1\\&=\;(m+1)&&+(m+1)&&\;+\cdots +(m+1)&&+(m+1)\quad {\text{ (}}n{\text{ terms)}}\\&=\;n\,(m+1)&&&&&&&&\\\end{alignedat}}}$$

so that ${\displaystyle S={\tfrac {1}{2}}\,n\,(m+1).}$ Substituting ${\displaystyle n={\tfrac {1}{2}}(m+1)}$ and ${\displaystyle m+1=2\,n}$ gives, respectively, the formulas

$${\displaystyle 1+3+\cdots +m\;=\;{\tfrac {1}{4}}(m+1)^{2}\quad {\text{ and }}\quad 1+3+\cdots +(2\,n-1)\;=\;n^{2}}$$

where the first formula expresses the sum entirely in terms of the odd integer ${\displaystyle m}$ while the second expresses it entirely in terms of ${\displaystyle n,}$ which is ${\displaystyle m}$'s ordinal position in the list of odd integers ${\displaystyle 1,3,5,\ldots .}$