Consider a Maclaurin series solution $$y = (1−x^2)y′′ −2xy′ +α(α+1)y=0, −1<x<1.$$
Show that $$a_2 = \frac{-α(α+1)}{6}a_0$$ $$a_3=\frac{−(α−1)(α+2)}{6}a_1$$
and, for all $n≥2$, $$a_{n+2} = \frac{n(n+1)−α(α+1)}{(n+2)(n+1)}a_n = \frac{(n−α)(n+α+1)}{(n+2)(n+1)}a_n.$$
Deduce that, if $α = k ∈ \{0,1,2,3,...\},$ then $$a_{k+2} =a_{k+4} =a_{k+6} =...=0.$$ Hence write down a polynomial solution of Legendre’s equation in the cases $α = 0,1,2,3,4$. For $α = 3$, write down the first 4 terms in the other series solution of Legendre’s equation and try to find a general expression for the coefficients in this series.
See my answer to another similar question: About the Legendre differential equation This gives you all the information you need.