$1 + 3 + 5 + . . . . + (2 n - 1) = n^{2}$ .

True

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False

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Solution

The correct option is A True
Let $P (n) : 1 + 3 + 5 + . . . . + (2 n - 1) = n^{2}$
Step 1 :
Put $n = 1$
Then, LHS = 1
RHS = $1^{2} = 1$
Therefore, LHS = RHS
$⟹ P (n)$ is true for $n = 1$ .

Step 2 :
Assume that $P (n)$ is true for $n = k$ .
Therefore, $1 + 3 + 5 + . . . . + (2 k - 1) = k^{2}$
Adding $2 k + 1$ on both sides, we get,
$1 + 3 + 5 + . . . . + (2 k - 1) + (2 k + 1) = k^{2} + (2 k + 1) = (k + 1)^{2}$
Therefore,
$1 + 3 + 5 + . . . . + (2 k - 1) + (2 (k + 1) - 1) = (k + 1)^{2}$
$⟹ P (n)$ is true for $n = k + 1$ .
Therefore, by the principle of mathematical induction P(n) is true for all natural numbers n.

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