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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In order to be parallel, the slope must be the same. You find the slope as the number or fraction connected to x. <em>("co-efficient" of x in math talk)</em>

In the given equation, that is -6. (So that knocks out the first two choices)

The other thing to look at is the y-value of the given coordinate,(-1,4)

<em>(The y-value is the second number in the coordinate (x.y) is the pattern)</em>

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The attachment shows what the graphs of the choices look like.

The black line is the correct answer. The given coordinate (-1,4) is the labeled red spot. The blue line is the given equation. (You can see where it "intercepts the y-axis on the +3) And the green line also has the -6 slope, but misses the point and intercepts the y-axis at 4.)

I hope the diagram and explanation helps you understand better. It can be confusing.