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Chapter 6: Problem 480

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

Expert verified
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

01

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.
02

Inductive hypothesis

Assume the statement P(k) is true for some natural number k: \(1 + 3 + 5 + \ldots + (2k-1) = k^2\)
03

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.
04

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

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