Trinomial Expansion

$(x+2y+3z)^{50} = [(x+2y)+3z]^{50} = \displaystyle \sum_{k=0}^{50} {\begin{pmatrix} 50 \\ k \end{pmatrix} (x+2y)^{50-k}(3z)^k}$

$(x+2y-3z)^{50} = [(x+2y)-3z]^{50} = \displaystyle \sum_{k=0}^{50} {(-1)^k \begin{pmatrix} 50 \\ k \end{pmatrix} (x+2y)^{50-k}(3z)^k}$

We note that the terms of $(x+2y+3z)^{50}$ and $(x+2y-3z)^{50}$ negate out when $k$ is odd. Therefore,

$(x+2y+3z)^{50} + (x+2y-3z)^{50} \\ = (x+2y)^{50} + \begin{pmatrix} 50 \\ 2 \end{pmatrix} (x+2y)^{48}(3z)^2 + \begin{pmatrix} 50 \\ 4\end{pmatrix} (x+2y)^{46}(3z)^4+...+(3z)^{50}$

Now, we note that: $(x+2y)^{50}$ has $51$ terms, $(x+2y)^{48}$ has $49$ terms and so on. Therefore, the total number of terms is:

$51+49+47+...+1 = \dfrac{26(1+51)}{2} = 26^2 = \boxed{676}$

Lets Assume x+2y = a Then On Seeing Expansion Alternate Terms Will get cancelled and we will be left with 26 terms.(Because the terms occupying even positions will get cancelled) now in these 26 terms we can replace a by x+2y The Power of a will vary like 50,48,46.......,2,0 in each of these there will be n+1 distinct terms so total number of distinct terms = 51+49+47.....+3+1 which is the sum of first 26 odd natural numbers and sum of first n odd natural numbers = n^2= 26^2 = 676