Question

If n is an odd positive integer, show that (n² - 1) divisible by 8<br /><br />pls Don't copy from Google <br />DO IT MANNUALY ..........PLSPLSPLSPLSPLSPLS​

Answer 1

Answer:

Step by step explanation:

Given: n is an odd positive integer.

To prove: that n² - 1 is divisible by 8.

Solution:

We know that,

Any Odd number is in the form of (2p +1) where p is a natural number.

n² - 1 = (2p + 1)² - 1

= 4p² + 4p + 1 -1

= 4p² + 4p

.°. n² - 1 = 4p² + 4p

Now, let's check by putting different values in place of p

p = 1,

4p² + 4p = 4(1)² + 4(1) = 4 + 4 = 8

Thus, it is divisible by 8.

p = 2,

4p² + 4p = 4(2)² + 4(2) =16 + 8 = 24

Thus, it is also divisible by 8.

p = 3,

4p² + 4p = 4(3)² + 4(3) = 36 + 12 = 48

Thus, it is also divisible by 8

Hence, we can conclude that 4p² + 4p is divisible by 8.

Proved that n² - 1 is divisible by 8.

Answer 2

Answer:

Given:

n is an odd positive integer.

To find:

(n² - 1) term is divisible by 8

Assumptions :

Since n is an odd positive integer, we can say that :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \large{ \red{{n = (4p + 1)}}}}$

here p is any positive whole number.

Calculation:

As per the question :

$= {n}^{2} - 1 \\ = {(4p + 1)}^{2} - 1$

$= \{16 {p}^{2} +( 2 \times 4p \times 1) + {1}^{2} \} - 1$

$= 16 {p}^{2} + 8p$

$= 8 \{2 {p}^{2} + p \}$

$= 8 \times k \: ......(divisible \: by \: 8)$

Let k be any other number.

So there is presence of 8 as a factor. Hence this term is divisible be 8.

$\boxed{ \large{ \bold{ \sf{ \red{hence \: proved}}}}}$

[Other wise , we can also put different values of p and check whether they are divisible by 8 or not.]