Question

Show that n2-n is divisible by 8 if n is an odd positive integer

Answer 1

Hello Dear.

There is the little Printing mistake in the Question. Correct Question would be like this ⇒
Show that (n² - 1) is divisible by 8 if n is an odd positive integer.

Solutions ⇒ 

To Prove ⇒ (n² - n) is divisible by 8.

Given ⇒ n is an odd positive integer.

Proof ⇒
We know that an Odd Positive Integer n is always written in the form of (4k + 1) or (4q + 3) or (4q + 5) and so on.

If n = 4k + 1 
Then, n² - 1 = (4k + 1)² - 1 
 = 16 k² + 1 + 8k - 1 
[∵ (a + b)² =  a²  + b² + 2ab]
 = 16k² + 8k
 = 8k(2k + 1)

Hence, it is divisible by 8.

If n = 4k + 3
Then, n² - 1 = (4k + 3)² - 1
 = 16k² + 9 + 24k - 1
 = 16k² + 24k + 8
 = 8(2k² + 3k + 1)


Hence, it is also divisible by 8.

If n = 4k + 5
Then, n² - 1 = (4k + 5) - 1
 = 16k² + 25 + 40k - 1
 = 16k² + 40k - 24 
 = 8(2k² + 5k - 3)

Hence, it is also divisible by 8.


Now, For any value of n, n² - 1 is always be divisible by 8.

Hence Proved.


Hope it helps.

Answer 2

Answer:

The answer is given in the picture pasted

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Step-by-step explanation: