I have a problem for Chebyshev polynomials. could you check whether my solution has fault or not
show by induction that $T_n(x)= \frac{(x+\sqrt {x^2-1})^n+(x-\sqrt{x^2-1})^n}{2}$
pf: when $n=0$, obviously $T_0(x) = 1$
Suppose $T_n(x)= \frac{(x+\sqrt {x^2-1})^n+(x-\sqrt{x^2-1})^n}{2}$ is ture for $n\ge 1$ then $$(x+\sqrt{x^2-1})T_n(x)=\frac{(x+\sqrt{x^2-1})^{n+1}+(x-\sqrt{x^2-1})^{n-1}}{2}$$
$$(x-\sqrt{x^2-1})T_n(x)=\frac{(x+\sqrt{x^2-1})^{n-1}+(x-\sqrt{x^2-1})^{n+1}}{2}$$
so$$2xT_n(x)=\frac{(x+\sqrt{x^2-1})^{n+1}+(x-\sqrt{x^2-1})^{n+1}}{2}+\frac{(x+\sqrt{x^2-1})^{n-1}+(x-\sqrt{x^2-1})^{n-1}}{2}$$
$$2xT_n(x)=\frac{(x+\sqrt{x^2-1})^{n+1}+(x-\sqrt{x^2-1})^{n+1}}{2}+T_{n-1}(x)$$
then we already know recursion formula of Chebyshev polynomials, that is $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$
So $$T_{n+1}(x)=\frac{(x+\sqrt{x^2-1})^{n+1}+(x-\sqrt{x^2-1})^{n+1}}{2}$$
Thus for all $n$ the proposition is true
I have some question about for using induction. When we suppose some proposition is true for $n \ge k$. Does it mean that we suppose the proposition is true for $n\lt k$? So can we use the condition for some $n \lt k$ when we prove for $n=k+1$? I use $\frac{(x+\sqrt{x^2-1})^{n-1}+(x-\sqrt{x^2-1})^{n-1}}{2}=T_{n-1}(x)$ above proof because we want to show only for $n=k+1$. Can it be possible like that? Thank you!
Yes, your proof looks good.
Regarding your question: this variant of induction where you not only suppose that the claim is true for the previous $n$, but for all $n<k$, is called complete induction (or "strong induction") and can be used for proving just like the usual induction (or "weak induction")