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Prove, using algebra, that $n(n^2 + 5)$ is even for all $n \in \mathbb{N}$. - Edexcel - A-Level Maths Pure - Question 13 - 2022 - Paper 2

Question 13

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Prove, using algebra, that $n(n^2 + 5)$ is even for all $n \in \mathbb{N}$.

Worked Solution & Example Answer:Prove, using algebra, that $n(n^2 + 5)$ is even for all $n \in \mathbb{N}$. - Edexcel - A-Level Maths Pure - Question 13 - 2022 - Paper 2

Step 1

Let $n = 2k$ (where $k \in \mathbb{N}$)

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Answer

When $n$ is even, we can substitute:

n(n^2 + 5) = (2k)((2k)^2 + 5) = (2k)(4k^2 + 5).

This expression is the product of an even number $2k$ and the sum $(4k^2 + 5)$ . Since $2k$ is even, the entire expression will be even regardless of whether $(4k^2 + 5)$ is even or odd.

Step 2

Let $n = 2k + 1$ (where $k \in \mathbb{N}$)

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When $n$ is odd, we can substitute:

n(n^2 + 5) = (2k + 1)((2k + 1)^2 + 5) = (2k + 1)(4k^2 + 4k + 1 + 5) = (2k + 1)(4k^2 + 4k + 6).

Here, $4k^2 + 4k + 6$ is even (as it is a sum of even numbers). We then have an odd number $(2k + 1)$ multiplied by an even number, which results in an even product.

Step 3

Conclusion

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Answer

In both cases, we have shown that whether $n$ is even or odd, the expression $n(n^2 + 5)$ is always even. Thus, we conclude that $n(n^2 + 5)$ is even for all $n \in \mathbb{N}$ .

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