Question

Q8. Use the principal of induction to prove that 1 + 3 + 5 + ... + (2n – 1) = n² for all n ∈ ℕ.​

Answer 1

$\large\underline{\sf{Solution-}}$

To prove that

$\rm \: 1 + 3 + 5 + \cdots \cdots + (2n - 1) = {n}^{2} \: \: \forall \: n \: \in \: N$

Let assume that

$\rm \: P(n): 1 + 3 + 5 + \cdots \cdots + (2n - 1) = {n}^{2}$

Step :- 1 For n = 1

$\rm \: P(1): 1 = {1}^{2}$

$\rm \: P(1): 1 = 1$

$\red{\rm\implies \:P(n) \: is \: true \: for \: n = 1 \: }$

Step :- 2 For n = k, Assume that P(n) is true, where k is some natural number.

$\rm \: P(k): 1 + 3 + 5 + \cdots \cdots + (2k - 1) = {k}^{2} - - (1)$

Step :- 3 For n = k + 1, we have to prove that P(n) is true.

That means, we have to prove that

$\rm \: P(k + 1): 1 + 3 + 5 + \cdots \cdots + (2k - 1) + (2k + 1)= {(k + 1)}^{2}$

Now, Consider

$\rm \: 1 + 3 + 5 + \cdots \cdots + (2k - 1) + (2k + 1)$

$\rm \: = \bigg(1 + 3 + 5 + \cdots \cdots + (2k - 1)\bigg) + (2k + 1)$

Now, using equation (1), we get

$\rm \: = \: {k}^{2} + 2k + 1$

$\rm \: = \: {(k + 1)}^{2}$

$\red{\rm\implies \:P(n) \: is \: true \: for \: n = k + 1 \: }$

Hence, By Principal of Mathematical Induction,

$\red{\rm \: 1 + 3 + 5 + \cdots \cdots + (2n - 1) = {n}^{2} \: \: \forall \: n \: \in \: N}$

Answer 2

To prove:-

We need to show two things:

1. Base case:

The equation holds true for n = 1

2. Inductive step: Assuming the equation holds true for n = k we need to show that it also holds true for n = k + 1

Let's proceed with the proof:

1. Base case:

n = 1

• Putting n = 1 into the equation we have:
• 1 = 1²
• This shows that the equation holds true for n = 1.

2. Inductive step:

Assume the equation holds true for n = k.

• That is we assume:
• 1 + 3 + 5 + ... + (2k - 1) = k² (Inductive Hypothesis)

Now,

Let's consider n = k + 1 :-

1 + 3 + 5 + ... + (2k - 1) + (2(k + 1) - 1)

We can simplify the left hand side of the equation using the inductive hypothesis:

k² + (2(k + 1) - 1)

Simplifying further:

k² + (2k + 2 - 1)

k² + (2k + 1)

Rearranging terms:

k² + 2k + 1

We can rewrite the right hand side of the equation as (k + 1)² :-

(k + 1)²

By the principle of induction if the equation holds true for n = k then it holds true for n = k + 1.

Therefore,

We have shown that:

1 + 3 + 5 + ... + (2n - 1) = n² for all n ∈ ℕ.

Therefore,

Te equation is proven using the principle of induction.