Prove $1 + 3 + 5 + \dots + (2 n - 1) = n^{2}$

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Solution

Let P(n): $1 + 3 + 5 + \dots + (2 n - 1) = n^{2}$ be the given statement
Step 1: Put $n = 1$
Then, L.H.S $= 1$
R.H.S $= 1^{2} = 1$
$∴$ L.H.S $=$ R.H.S
$⟹$ P(n) is true for $n = 1$
Step 2: Assume_that P(n) is true for $n = k$
$∴ 1 + 3 + 5 + \dots + (2 k - 1) = k^{2}$
Adding $2 k + 1$ on both sides, we get
$1 + 3 + 5 + \dots + (2 k - 1) + (2 k + 1) = k^{2} + (2 k + 1) = (k + 1)^{2}$
$∴ 1 + 3 + 5 + \dots + (2 k - 1) + (2 (k + 1) - 1) = (k + 1)^{2}$
$⟹$ P(n) is true for $n = k + 1$
$∴$ By the principle of mathematical induction P(n) is true for all natural numbers 'n'.
Hence $1 + 3 + 5 + \dots + (2 n - 1) = n^{2}$ , for all $n \in N$

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