Question

If \(\mathrm{P}(\mathrm{n}): 1+3+5+\ldots \ldots+(2 \mathrm{n}-1)=\mathrm{n}^{2}\) is (a) True for \(\mathrm{n}>1\) (b) True for all \(\mathrm{n} \in \mathrm{N}\) (c) True for no \(\mathrm{n}\) (d) None of these

Short answer

The given statement P(n) is true for the base case of n=1 and also holds true for all natural numbers by the method of induction. Therefore, the correct answer is (b) True for all n ∈ ℕ.

Step-by-step solution

Verify the formula for the base case

Choose the smallest value for which the statement needs to be checked. In this case, we need to verify it for n=1 (as it belongs to natural numbers). Sum of first n odd numbers: \(1 = 1^2\) Since the statement holds true for n=1, we can move on to the next step.

Inductive hypothesis

Assume the statement P(k) is true for some natural number k: \(1 + 3 + 5 + \ldots + (2k-1) = k^2\)

Inductive step

Now, we need to prove P(k+1) is true: \(1 + 3 + 5 + \ldots + (2k-1) + (2(k+1)-1) = (k+1)^2\) If we replace P(k) in the above expression, we get: \(k^2 + (2(k+1) - 1) = (k+1)^2\) By simplification, we get: \(k^2 + 2k + 1 = k^2 + 2k + 1\) Since both sides of the equation are equal, P(k+1) is true.

Conclusion

Since the statement holds true for the base case, and P(k+1) is true when P(k) is assumed to be true, we can conclude the given formula is true for all natural numbers (n). So, the correct option is: (b) True for all n ∈ ℕ

Key concepts

Odd Numbers Sum

The sum of the first few odd numbers forms an interesting sequence in mathematics. Specifically, the sum of the first \( n \) odd numbers is equal to \( n^2 \). This is quite an intriguing pattern and is represented by the mathematical expression: \( 1 + 3 + 5 + \, ... \, + (2n-1) = n^2 \).

• For \( n=1 \), the sum is \( 1 \).
• For \( n=2 \), the sum is \( 1+3=4 \), which is \( 2^2 \).
• For \( n=3 \), the sum is \( 1+3+5=9 \), which is \( 3^2 \).

This consistent pattern allows us to predict that for any natural number \( n \), the sum of its first \( n \) odd numbers will equal \( n^2 \). Recognizing patterns like these helps students understand deeper mathematical concepts and builds a solid foundation for using mathematical induction.

Inductive Hypothesis

Inductive hypothesis is an essential part of mathematical induction. It's a method where you assume a statement is true for some arbitrary number \( k \), and then use this assumption to prove the statement is true for the next number, \( k+1 \). In our exercise, the inductive hypothesis is assuming the formula \( 1 + 3 + 5 + \ldots + (2k - 1) = k^2 \) holds true for some natural number \( k \).

• This assumption acts as a bridge between the base case and proving the next step.
• It simplifies complex mathematical proofs by breaking them into smaller, manageable parts.

Using this hypothesis is crucial, as it is the foundation upon which the inductive step is built, ultimately proving the overall statement for all natural numbers.

Inductive Step

After setting the inductive hypothesis, the inductive step is where we prove that if the statement is true for \( k \), it must also be true for \( k+1 \). This involves replacing the formula with the assumption we've made for \( k \), and then logically extending it to \( k+1 \).In our exercise:

• We assume \( 1 + 3 + 5 + \ldots + (2k - 1) = k^2 \) is true.
• Next, we calculate \( 1 + 3 + 5 + \ldots + (2k - 1) + (2(k+1) - 1) \), which should equal \( (k+1)^2 \).
• Upon simplifying, both sides equal \( k^2 + 2k + 1 \), confirming \( (k+1)^2 \).

This proof ensures that if the formula holds for any natural number \( k \), it will hold for \( k+1 \) as well, thus establishing the statement's truth for all natural numbers.

Base Case Verification

Mathematical induction begins with a base case verification. This is when you verify the mathematical statement for the initial value of the variable, which in most cases is \( n=1 \). It sets the groundwork for the entire proof to be considered valid. For our exercise, the base case is checking if the formula is true when \( n=1 \).

• The left-hand side becomes \( 1 \).
• The right-hand side, \( n^2 \), becomes \( 1^2 = 1 \).

Since both sides are equal, it confirms the base case is true. This base case is crucial because if the original equation does not hold for the base case \( n=1 \), the entire process of mathematical induction cannot proceed. Once the base case is verified, it clears the path for the inductive hypothesis and step to prove the statement for all natural numbers.