If $X,Y\sim B$ so $X+Y\sim B$? Let $X,Y$ be random independent variables, such that $X\sim B(n,0.5),Y\sim B(n,0.2)$


  $X+Y$ is also binomial distributed?


My start:

$$X+Y\in \{0,1,2,3,\dots\}$$

$$P(X=k)=\binom{n}{k} 0.5^k(1-0.5)^{n-k}=\binom{n}{k} 0.5^n$$ 

$$P(X=t)=\binom{n}{t} 0.5^t(1-0.5)^{n-t}=\binom{n}{t}0.5^n$$ 

Let as denote $S:=X+Y\,\,\,\,\,\,\,\,\,\,\,\,\,\,$
$P(S=\mathscr{L})$

I'm stuck here
It is not binomial distributed. In general though there is a very reliable method so establish if a linear combination of independent variables follows some distribution: compute their charachteristic function!

So in this case 
$$E[e^{iu(X+Y)}] = E[e^{iuX}]E[e^{iuY}]  = ((1-p_1) + p_1e^{iu})^n((1 - p_2) + p_2e^{iu})^n$$ and with some algebra you can readily see that this is not in the form $((1-p) + pe^{iu})^n$
I think you are stuck because the result you are trying to show isn't quite right. To the best of my knowledge, there is no named distribution that $X + Y$ follows, and it most definitely is not binomial. 

As a simple illustration why, simply consider the $n=1$ case, which has PMF $P(X+Y=0) = 0.4$, $P(X + Y = 1) = 0.35$, $P(X + Y= 2) = 0.25$. If this were binomial, $n$ would have to equal 2 because the support of this distribution is $\{0,1,2\}$ and $p$ would have to equal $0.5$ because $P(X + Y=2) = p^2 \implies p = 0.5$. However, if we looked at $P(X+Y=0) = 1-p^2 \implies p = \sqrt{0.6}$ which draws a contradiction.

The only case where summing binomials gives you another binomial is when the probabilities are equal, thus if $X_1 \sim B(n_1, p)$ and $X_2 \sim B(n_2, p)$ then $X_1 + X_2 \sim B(n_1 + n_2, p)$.