Using the fact that $1+2+\cdots+n=\frac{n(n+1)}{2}$, we can deduce that sum of first $n$ positive odd integers is $n^2$. However, is there a way of finding the sum of $1+3+5+\cdots+(2n-1)$ visually?
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6$\begingroup$ en.wikipedia.org/wiki/Proof_without_words#Sum_of_odd_numbers $\endgroup$– Jaap ScherphuisCommented Aug 10, 2020 at 14:34
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3$\begingroup$ youtube.com/watch?v=IJ0EQCkJCTc $\endgroup$– abcd123Commented Aug 10, 2020 at 14:39
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4$\begingroup$ math.stackexchange.com/a/733805/179297 $\endgroup$– JMoravitzCommented Aug 10, 2020 at 14:47
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1$\begingroup$ $\displaystyle \begin{array}{ccc} \color{red}{\huge\bullet} & \color{magenta}{\huge\bullet} & \color{black}{\huge\bullet} \\ \color{magenta}{\huge\bullet} & \color{magenta}{\huge\bullet} & \color{black}{\huge\bullet} \\ \color{black}{\huge\bullet} & \color{black}{\huge\bullet} & \color{black}{\huge\bullet} \end{array} $ $\endgroup$– Felix MarinCommented Aug 11, 2020 at 5:19
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1 Answer
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Here is a ‘proof’ I once found in a book for young children. It is not a real proof in the mathematical sense, but rather a convincing example that any mathematician feels could be transformed into a rigourous proof:
Imagine wooden cubes stacked in rows, with the basis containing, say, $7$ cubes, the row above, $5$ cubes, the row still above, $3$ and the last row $1$, like this:
It is a geometrical evidence that, moving the grey squares from the bottom right corner to the top left corner, one recreates a square with sides equal to the number of rows, i.e. $4$ units, hence we have $16$ of them for the sum of the $4$ first odd numbers.
