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that the sum of the `\(n^{th}\) first odd numbers is equal to \(n^2\).
A odd number \(x\) can be noted \(2k-1\). We are looking at \(\sum_{i=1}^{n}(2i - 1)\).
And we have \(\sum_{i=1}^{n}(-1) = -n\).
So \(\sum_{i=1}^{n}(2i - 1) = 2\left(\sum_{i=1}^{n}i\right) - n\)
\(= 2 \times \frac{n^2 + n}{2} - n = n^2\).

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