I understand the concept just fine, it's the more complicated arithmetic that I struggle with.

Like my homework came down to showing that (1/2 * k * (k+1)) ^2 + (k+1) ^ 3 == (1/2 * (k+1) * (k+2))^2 and I just don't even know what to do. Another one was to show that (k+1)^3 + 5(k+1) == 6j for some integer j>= 1.

Thanks if you can help.

I try to help abit, only on phone & long time ago since I last did it^^

Could give more precise help tomorrow when I have a comp and more time:

Start of with showing that your equation is right for 1 integer, e.g. for 2 for the 2nd example - this is called anchor (in germany)

Than take your equation and put k+1 into it. Rearange it to get the equation for k (which you showed to be true with the anchor). The remaining part that is the "+1" to your k. And done.

Hope that helped, check wikipedia too maybe. These are common examples for induction

You want to prove something true, that depends on a given integer k.
Let P(k) the sentence you are working with. For each integer k, P(k) could be true or false.

Let's start with k = 0. Check if P(0) is true. This is commonly an easy job.

Now assume that, for some given k ≥ 0, P(k) is true. And try to prove, in this case, that P(k+1) is also true.
If you succeed, then it's safe to say that P(k) is true for every integer k ≥ 0.

Example with your second query :

P(k) is " the number (k+1)^3 + 5(k+1) is a multiple of 6 ".

P(0) is " the number 6 is a multiple of 6 ".
P(0) is true.

Let k ≥ 0, assume that P(k) is true.

Let's try to prove P(k+1) true.
P(k+1) tells something about the number (k+2)^3 + 5(k+2). And we already now something about the number (k+1)^3 + 5(k+1).

Let's try to make the second appear from the first :
(k+2)^3 + 5(k+2) = ( k+1 +1)^3 + 5(k+1) + 5
= (k+1)^3 + 3.(k+1)^2 + 3.(k+1) + 1 + 5(k+1) + 5
= (k+1)^3 + 5(k+1) + 3(k+1)(k+1+1) + 6
= (k+1)^3 + 5(k+1) + 3(k+1)(k+2) + 6

(k+1)(k+2) is an even integer (since one among k+1 or k+2 must be even), hence 3(k+1)(k+2) is a multiple of 6.

And since you have assumed that P(k) is true, we can say that (k+1)^3 + 5(k+1) is a multiple of 6.

Finally, (k+2)^3 + 5(k+2) is a sum of multiples of 6, hence P(k+1) is also true.