Show that n^2-1 is divisible by 8,if 'n' is an odd positive integer.

1 answer:

6 0

$n-an\ odd\ positive\ integer\ \ \ \Rightarrow\ \ \ n=2k+1\ \ \ and\ \ \ k\in Natural\\\\n^2-1=(2k+1)^2-1=(2k)^2+2\cdot2k\cdot1+1^2-1=4k^2+4k=\\\\=4k(k+1)\\\\if\ k-odd\ \ then\ (k+1)-even,\ \ \ then\ k(k+1)-even\\\\\ \ \ \Rightarrow\ \ \ k(k+1)=2m\ \ \ and\ \ \ m\in N\ \ \ \Rightarrow\ \ \ 4k(k+1)=4\cdot2m=8m\\\\\\if \ k-even,\ \ \ then\ k(k+1)-even\\\\\ \ \Rightarrow\ \ \ k(k+1)=2p\ \ \ and\ \ \ p\in N\ \ \ \Rightarrow\ \ \ 4k(k+1)=4\cdot2p=8p\\\\$

$------------------------\\\\ for\ each\ k \in N\ the\ product\ of\ \ \ k (k +1)\ \ \ is\ even,\ so:\\\\if\ 'n'\ is\ an \ odd\ positive\ integer,\ then\ \ \ n^2-1\ \ \ is\ divisible\ 8.$

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Rewriting:
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Answer:

The lengths of the sides of the square base are $2\sqrt{2}\ units$

Step-by-step explanation:

The correct question is

A triangular prism has an isosceles right triangular base with a hypotenuse of square root of 32 and a prism height of 12. a square prism has a height of 12 and its volume is equal to that of the triangular prism. what are the dimensions of the square base, in simplest radical form?

we know that

The volume of a prism is given by the formula

$V=Bh$

where

B is the area of the base

h is the height of the prism

If the volumes of the solids are the same and their heights are the same, then the area of the base of the triangular prism must be equal to the area of the base of the square prism

step 1

Find the area of the base of the triangular prism

Remember that in a right isosceles triangle, the two legs are equal

Let

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Applying the Pythagorean Theorem

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