if we let x= (y +1/y) , and call

yn+ 1/yn = Pn (the n should be suffices throughout)

so that P1 -y +1/y = x

then we can find P2, P3, etc and ideally derive a formula for Pn in terms of x?

for example (y +1/y)²= P2 => P2= x²=2

similarly P3= x³- 3y

I can get P4- P10 but can’t get to a formula

Can anyone help?

I think a lot of this is garbled (including the word "suffices", which is probably meant to be "superscripts" or "exponents"). It will be good if you can show us an image of the source from which you got this idea, and/or of your work on paper, so we can be sure what you mean.

I think it may be something like this:

If we let x = y + 1/y, that is, `x = y + 1/y`,

and P_n = y^n + 1/y^n, that is, `P_n = y^n + 1/y^n`,

so that P_1 = y + 1/y = x, that is, `P_1 = y + 1/y = x`,

then we can successively find P_2, P_3, and so on to derive a formula for P_n in terms of x.

For example, x^2 = (y + 1/y)^2 = y^2 + 2 + 1/y^2 = P_2 + 2, that is, `x^2 = (y + 1/y)^2 = y^2 + 2 + 1/y^2 = P_2 + 2`

so that P_2 = x^2 - 2, that is, `P_2 = x^2 - 2`.

Part of what I am doing is teaching you a better notation for typing math; and everything that follows "that is" is the very same thing put between "back quotes", which on my keyboard is just below ESC. That provides an easy way to display math neatly.

Let me know if I am interpreting correctly, and then show us your work (perhaps as an image) so we can see whether you did it correctly, and then work on bringing Pascal's triangle into this and maybe finding a general formula.