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3 years ago

Prove that (4n+1)^2(4n-1)^2 is a multiple of 8 for all integers of n. thanks.

Mathematics

1 answer:

azamat3 years ago

6 0

Answer: i think (4n+1)^2(4n-1)^2 isnt a multiple of 8 for all integers of n because:

(4n + 1)²(4n - 1)²

= [(4n + 1)(4n - 1)]²

= (16n² - 1)²

= 16².n².n² - 2.16.n² + 1

= 8n²(32n² - 4) + 1

can see 8n²(32n² - 4) is a multiple of 8 but 1 isnt a multiple of 8

=> (4n + 1)²(4n - 1)² isnt a multiple of 8 for all integers of n.

Step-by-step explanation:

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Now that we have the number of tickets that the adults bought, lets replace that value in equation (2):
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3 years ago

Read 2 more answers

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Answer:

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Step-by-step explanation:

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Answer:

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