Q1 a) Prove that the sum of the squares of the first n integers is 1/24 (2n) (2n+1) (2n+2)?

a/. Let's prove this using induction. First show it to be true for n=1. The sum of the first 1 integer squared is 1² = 1. Substituting n = 1 into

1/24(2n)(2n+1)(2n+2) gives 1/24(2)(3)(4), which is indeed equal to 24/24 = 1.

Now assume it is true for n=k. In other words, you need to assume the sum of the squares of the first k integers is 1/24(2k)(2k+1)(2k+2). You need to use this assumption to show it is also true for n=k+1.

The sum of the squares of the first k+1 integers is equal to the sum of the first k integers plus (k+1)².

So it is equal to

[1/24 (2k)(2k+1)(2k+2)] + (k+1)²

= [1/24 (2k)(2k+1)(2k+2)] + k² + 2k + 1

= [1/24 (2k)(2k+1)(2k+2)] + 24(k² + 2k + 1)/24

= [(2k)(2k+1)(2k+2) + 24(k²+2k+1)]/24

Expand out the (2k)(2k+1)(2k+2) part to get (4k²+2k)(2k+2) = 8k³+8k²+4k²+4k = 8k³+12k²+4k . So we have

= [(2k)(2k+1)(2k+2) + 24(k²+2k+1)]/24

= [8k³+12k²+4k + 24(k²+2k+1)]/24

= [8k³+12k²+4k + 24k²+48k+24)]/24

= [8k³+36k²+52k+24]/24

This is the sum of the squares of the first k+1 integers. We want to show this is equal to

1/24(2(k+1))(2(k+1)+1)(2(k+1)+2)

=1/24(2k+2)(2k+3)(2k+4)

= (2k+2)(2k+3)(2k+4)/24 and expand out the brackets

= (4k²+10k+6)(2k+4)/24 and expand out the remaining brackets

= [8k³+16k²+20k²+40k+12k+24]/24 and simplify by adding like terms to get

= [8k³+36k²+52k+24]/24

We see indeed that they are equal, hence it is true.

(b) A prime number must be a positive integer. All positive integers have either a remainder of 0, 1 or 2 when divided by 3. Since p>= 5, it cannot be 3. And if it is prime, it cannot be divisible by 3. A number is divisible by 3 if and only if it has a remainder of 0 when divided by 3. So it cannot have a remainder of 0 when divided by 3 and therefore must have a remainder of 1 when divided by 3.

If p has a remainder of 1 when divided by 3, then there is some integer m such that p = 3m+1. And therefore p²+2 = (3m+1)²+2 = 9m²+6m+1+2 = 9m²+6m+3 = 3(3m²+2m+1) so it will always be divisible by 3 if p has a remainder of 1 when divided by 3.

If p has a remainder of 2 when divided by 3, there is some integer a such that p = 3a+2. And therefore p²+2 = (3a+2)² = 9a²+12a+4+2 = 9a²+12a+6 = 3(3a²+4a+12) so it will always be divisible by 2 if p has a remainder of 2 when divided by 3.

So if p is a prime greater than or equal to 5, p²+2 will always be divisible by 3 and therefore composite.

Multiply everything by (n - 1)(n - 2) to cancel out all denominators: 2n(n - 2) - 2(n - 1) = (n - 1)(n - 2) Expand the brackets: 2n^2 - 4n - 2n + 2 = n^2 - 3n + 2 Combine all the like terms: 2n^2 - 6n + 2 = n^2 - 3n + 2 Move everything to one side: n^2 - 3n = 0 Factor out n: n(n - 3) = 0 n = 0, 3