I was thinking about excercise 9-A in Milnor's Characteristic classes.
We denote $\gamma^n$ to be an $n$-plane bundle over the infinite Grassmanian $G_n(\mathbb{R}^\infty)$. We want to show that if $\gamma^n\oplus \gamma^n$ is an orientable bundle that $w_{2n}(\gamma^n\oplus\gamma^n)\neq 0$. My question is how do we know that $$w_{2n}(\gamma^n\oplus\gamma^n)=\Sigma_{i+j=n} w_i(\gamma^n)\cup w_j(\gamma^n),$$ I mean, using the Whitney formula, the LHS is expressed on the right, only on the top rank $n$?
I'm not sure what you're asking exactly, but Whitney formula in this context states that, $$ w_{2n}(\gamma^n \oplus \gamma^n) = \sum_{i+j=2n} w_i(\gamma^n) \cup w^j(\gamma^n). $$ Note we are summing over $i+j = 2n,$ and not $n.$
To proceed, observe that $w_i(\gamma^n) = 0$ for all $i > n,$ since $\gamma^n$ is $n$-dimensional. Hence the only non-zero term corresponds to $i = j = n.$ That is, $$ w_{2n}(\gamma^n \oplus \gamma^n) = w_n(\gamma^n)^2 \in H^{2n}(G_n(\mathbb R^{\infty});\mathbb Z/2). $$ Since $H^{*}(G_n(\mathbb R^{\infty});\mathbb Z/2)$ is freely generated by $w_i(\gamma^n)$ for $1\leq i \leq n,$ it follows that $w_{2n}(\gamma^n \oplus \gamma^n) \neq 0.$