$$(1+2x+3x^2)^{50}=a_{0}+ a_{1}x +...+ a_{100}x^{100}$$
Find ratio between $a_{51}$ and $a_{49}$. Another sub question asked is to find the relation between $a_{n}, a_{n-1}, a_{n-2}$
My approach
I tried to apply the standard binomial theorem that is $(1+t)^n$ keeping $t$ as $2x+3x^2$ but that made it too calculative. How else can we approach sub problems with greater than 2 terms.
Write $ x = \frac{1}{3z} $ in the equation and then multiply both sides by $ (3z^2)^{50} $. The left hand side becomes $ (1+2z+3z^2)^{50} $ and the right hand side becomes $ \sum_{j=0}^{100} a_j3^{50-j}z^{100-j} $. But by definition, the new left hand side is also $ \sum_{k=0}^{100} a_kz^k $. This gives the relation $ a_j3^{50-j} = a_{100-j} $ for every $ j $. In particular we find that $ 3a_{49} = a_{51} $.