Recent content by silina01

I found the answer, thank you all so much!

okay, but I am still stuck

Define the sequence of integers a1, a2, a3,... as follows: a1 = 3 a2 = 6 an = 5an-1 - 6an-2 + 2 for all n ≥ 3 Prove that an = 1 + 2n-1 + 3n-1 ------------------------------------------------------------------------------------------------ my attempt: base case: n=1 1+ 20 +30 = 1...

x and y cannot be 0 by the way

what would that be?

ahhhh I see it know, in both cases if x-y <0 or if x-y>0 the quotient will always be positive. Thanks everyone.

(x^5 - y^5)/(x - y) = x^4 + x^3y + x^2y^2 + xy^3 + y^4 , is that correct?

x^n - y^n = (x - y) (x^n-1 + x^n-2y+ ...+ xy^n-2 + y^n-1 )??

I am still lost

I factored out x^4 and got x^4 [(y/x)^4 + (y/x)^3 +(y/x)^2 + y/x +1] I see that all these terms have y/x so if i let t = y/x then it'll be x^4 [t^4 + t^3 +t^2 + t +1] (just so it is easier to visualize) but I am stuck, How do i simplify it further?

sorry I meant to say that if x and y are both not 0

Prove that if x and y are not both , then x^4+x^3y+x^2y^2+xy^3+y^4 > 0 I have no idea how to start this proof, can anyone give me an idea?