15. (i) Use proof by exhaustion to show that for $n \in \mathbb{N}, n 3^n$ (ii) Given that $m^2 + 5$ is odd, use proof by contradiction to show, using algebra, that $m$ is even.

A-Level Edexcel Maths Pure Functions: 15. (i) Use proof by exhaustion

15. (i) Use proof by exhaustion to show that for $n \in \mathbb{N}, n < 4$ $(n + 1)^3 > 3^n$ (ii) Given that $m^2 + 5$ is odd, use proof by contradiction to show, using algebra, that $m$ is even. - Edexcel - A-Level Maths Pure - Question 2 - 2021 - Paper 2

Question 2

$15.-(i)-Use-proof-by-exhaustion-to-show-that-for-$n-\in-\mathbb{N},-n-<-4$---$(n-+-1)^3->-3^n$---(ii)-Given-that-$m^2-+-5$-is-odd,-use-proof-by-contradiction-to-show,-using-algebra,-that-$m$-is-even.-Edexcel-A-Level Maths Pure-Question 2-2021-Paper 2.png$

15. (i) Use proof by exhaustion to show that for $n \in \mathbb{N}, n < 4$ $(n + 1)^3 > 3^n$ (ii) Given that $m^2 + 5$ is odd, use proof by contradiction to show... show full transcript

Worked Solution & Example Answer:15. (i) Use proof by exhaustion to show that for $n \in \mathbb{N}, n < 4$ $(n + 1)^3 > 3^n$ (ii) Given that $m^2 + 5$ is odd, use proof by contradiction to show, using algebra, that $m$ is even. - Edexcel - A-Level Maths Pure - Question 2 - 2021 - Paper 2

Step 1

Use proof by exhaustion to show that for $n \in \mathbb{N}, n < 4$ $(n + 1)^3 > 3^n$

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Answer

To prove this statement, we will evaluate $(n + 1)^3$ and $3^n$ for all natural numbers $n$ that are less than 4:

For $n = 1$ :
- $(1 + 1)^3 = 2^3 = 8$ and $3^1 = 3$ , hence $8 > 3$ .
For $n = 2$ :
- $(2 + 1)^3 = 3^3 = 27$ and $3^2 = 9$ , hence $27 > 9$ .
For $n = 3$ :
- $(3 + 1)^3 = 4^3 = 64$ and $3^3 = 27$ , hence $64 > 27$ .

Since the inequality holds for all values of $n = 1, 2, 3$ , we conclude using proof by exhaustion that for $n \in \mathbb{N}, n < 4$ , $(n + 1)^3 > 3^n$ .

Thus the statement is proved.

Step 2

Given that $m^2 + 5$ is odd, use proof by contradiction to show, using algebra, that $m$ is even.

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Answer

Assume, for the sake of contradiction, that $m$ is odd. Then we can express $m$ as:

$m = 2k + 1$

for some integer $k$ . Therefore:

$m^2 = (2k + 1)^2 = 4k^2 + 4k + 1$

Substituting this into the expression for $m^2 + 5$ gives:

$m^2 + 5 = (4k^2 + 4k + 1) + 5 = 4k^2 + 4k + 6$

Notice that $4k^2 + 4k + 6$ is clearly even, since all terms are multiples of 2. Thus, we find:

$m^2 + 5$ is even, which contradicts the assumption that $m^2 + 5$ is odd. Therefore, our assumption must be incorrect, and we conclude that $m$ must be even.

15. (i) Use proof by exhaustion to show that for $n \in \mathbb{N}, n < 4$ $(n + 1)^3 > 3^n$ (ii) Given that $m^2 + 5$ is odd, use proof by contradiction to show, using algebra, that $m$ is even. - Edexcel - A-Level Maths Pure - Question 2 - 2021 - Paper 2

Worked Solution & Example Answer:15. (i) Use proof by exhaustion to show that for $n \in \mathbb{N}, n < 4$ $(n + 1)^3 > 3^n$ (ii) Given that $m^2 + 5$ is odd, use proof by contradiction to show, using algebra, that $m$ is even. - Edexcel - A-Level Maths Pure - Question 2 - 2021 - Paper 2

Use proof by exhaustion to show that for $n \in \mathbb{N}, n < 4$ $(n + 1)^3 > 3^n$

Given that $m^2 + 5$ is odd, use proof by contradiction to show, using algebra, that $m$ is even.

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