Prove by mathematical induction
\[
\sum_{i=1}^n (2i-1) = n^2
\]
for all natural numbers n > 0.
Prove the statement is true for a particular n. Here n = 1 looks a reasonable choice.
To show this is true for n=1, observe that the left hand side = 1 and the right hand side = 1.
So the statement is true for n=1.
Assume the statement is true for n = k and show it is true for n = k + 1.
To show the statement is true for n = k+1, call the LHS \( S_{k+1} \). Then
\[ \begin{align}
S_{k+1} &= S_k + 2(k+1) - 1 \\
\end{align} \ \]
Use the inductive hypothesis to simplify the right hand side.
By the inductive hypothesis,
\[ \begin{align}
S_{k+1} &= \sum_{i=1}^k (2i-1) + 2(k+1) - 1 \\
&= k^2 + 2(k+1) - 1 \\
&= k^2 + 2k + 1 \\
&= (k+1)^2
\end{align} \ \]
Now the RHS has the same form as the RHS of the original expression with k+1 substituted for n. So the statement is true for n = k+1.
By the principle of mathematical induction,
The statement is true for \( n \gt 0 \).
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