This is based on distributive law where k(k-1)^n - (k-1)^n = (k-1)^1*(k-1)^n = (k-1)^(n+1).
This is just based on ab + cb = (a+c)*b.
Hello, I am having a hard time understanding why k(k - 1)^n - (k - 1)^n - (-1)^n(k - 1) = (k - 1)^(n+1) + (-1)^(n+1) (k - 1).
What i specifically don't get is how to get from ^n to ^n+1. I would appreciate it if someone could tell me how this is done.
Let b = (k-1)^n, a = k and c = -1.
We know from distributive law that (a+c)b = ab + bc if we expand out but we can do the reverse: if we have ab + bc we can write this as (a+c)b.
Now ab = k*(k-1)^n and bc = -(k-1)^n so (a+c)b = (k + -1)*(k-1)^n = (k-1)*(k-1)^n but from powers of terms, we have z^a * z^b = z^(a+b) (exponentials) so in this case z = (k-1), a = 1 and b = n we get a+b = n+1 with our z so the final result is (k-1)^(n+1).
Ok so your term is - (-1)^(n+1). Now this is the same as (-1)^n * (-1). But this changes your - (since this is there before the term) to a plus if you expand it out.
So your - (-1)^n * (-1) changes the - to a + giving - - (-1)^n = + (-1)^n.
I understand that x^n * 1^n = x^n+n, but i just don't understand how this logic is being applied to the polynomial in the original post.
actually, on rereading your original post, chiro, i think i get what you are saying. So you are manipulating the polynomial using the distribute law. I think i need to read up on this law. I know about it from logic, but i'm a bat hazy at applying it to numbers