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