Some help with maths please? 2 'proof' questions!?

Question

Thanks! I am really stuck on proof and I was wondering if I could have some help? One question is the 'nth' even number is 2n. The next even number after 2n is 2n+2 - Explain why.

The second is prove that (3n+1)squared - (3n-1)squared is a multiple of 4, for all positive integer values of n.

Thank you so much for help in advance! I would ask my teacher but it is in for monday first thing!

Answer 1

because even numbers are multiples of 2, so 2n is, and 2n+2 is the next number that can be a multiple of 2.... this question is lame, i am sorry





a² + b² = (a + b) (a - b) >>>> a = (3n+1) and b = (3n-1)

(a+b) = [(3n+1) + (3n-1)] , (a-b) = [(3n+1) - (3n-1)]

(3n+1)² - (3n-1)² = (3n+1+3n-1) (3n+1 - 3n +1)

(3n+1)² - (3n-1)² = (6n) (2)

(3n+1)² - (3n-1)² = 12 n since 12 is a multiple of 4, whatever integer n equals to the product will be a multiple of 4.

Answer 2

(3n + 1)^2 - (3n - 1)^2



= 9n^2 + 6n + 1 - 9n^2 + 6n - 1



= 12n



= 4(3n)



hence 4 is a factor for all positive integer values of n

Answer 3

Let x, y be truly numbers with xzero , WHY????????) Assume that xzero and permit y'=y+okay Then given that xzero. So we will be able to use x' and y' as a substitute of x and y for the evidence. That's what is supposed through no loss in generality. So what do we now have? zero x' + one million/m and x' >= (n-one million)/m then y' > (n-one million)/m + one million/m or y'>n/m So we now have y'>n/m>x'>(n-one million)/m which finishes the evidence.