Prove by induction:
$1 + 3 + 5 + . . . . . + (2 n - 1) = n^{2}$

Open in App

Solution

The statement to be proved is:
$P (n) = 1 + 3 + 5 + . . . + (2 n - 1) = n^{2}$

Step 1: Veriy that $P (1)$ is correct:
$P (1) : 2 (1) - 1 = 1^{2} P (1) : 1 = 1$
Therefore it is verified that $P (1)$ is correct.

Step 2: Assume that $P (k)$ is true
Let us assume that:
$P (k) : 1 + 3 + 5 + . . . + (2 k - 1) = k^{2}$ holds.

Step 3: Prove that $P (k + 1)$ is true.
$L H S = 1 + 3 + 5 + . . . + (2 (k + 1) - 1)$
$= P (k) + (2 k + 1)$
$= k^{2} + 2 k + 1$
$= (k + 1)^{2}$
$= R H S$
Hence, $P (n)$ is true by Principle of Mathematical Induction

Suggest Corrections

Similar questions

\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2 n - 1) (2 n + 1)} = \frac{n}{2 n + 1}

n \in N

(1 + \frac{3}{1}) (1 + \frac{5}{4}) (1 + \frac{7}{9}) \dots (1 + \frac{(2 n + 1)}{n^{2}}) = (n + 1)^{2}

Prove the following by using the principle of mathematical induction for all n ∈ N:

n \in N : (1 + \frac{3}{1}) (1 + \frac{5}{4}) (1 + \frac{7}{9}) . . . . . . (1 + \frac{(2 n + 1)}{n^{2}}) = {(n + 1)}^{2}

1^{2} + 2^{2} + 3^{2} + . . . . . . . . + n^{2} = \frac{1}{6} n (n + 1) (2 n + 1)

Join BYJU'S Learning Program