Let $a,b\in\mathbb{F}_{2^{m}}$ (a field of characteristic $2$, m is odd) I need to prove that
$B(a,b)=tr(\displaystyle\sum_{i=1}^{(m-1)/2}(a+b)^{1+2^{i}})-tr(\displaystyle\sum_{i=1}^{(m-1)/2}a^{1+2^{i}})-tr(\displaystyle\sum_{i=1}^{(m-1)/2}b^{1+2^{i}}),$
where $tr:\mathbb{F}_{2^{m}}\rightarrow\mathbb{F}_2$ is the trace function, is a bilinear form with full rank.
Correcting the argument to reflect the new summation range. The freshman's dream implies (as you had apparently figured out in the alternate version of this question) that for all $i$ and all $a,b$ we have $$ (a+b)^{1+2^i}-a^{1+2^i}-b^{1+2^i}=a^{2^i}b+ab^{2^i}. $$ Let $m=2k+1$, so $(m-1)/2=k$. We shall need the fact that conjugate elements have the same trace: $tr(x)=tr(x^2)$. Iterating this relation $m-i$ times gives $$ tr(ab^{2^i})=tr(a^{2^{m-i}}(b^{2^i})^{2^{m-i}})=tr(a^{2^{m-i}}b^{2^m})=tr(a^{2^{m-i}}b). $$ Therefore the three summations (forget the trace here temporarily) can be combined to read $$ \sum_{i=1}^k(a^{2^i}b+ab^{2^i})=\left(\sum_{i=1}^k(a^{2^i}+a^{2^{m-i}})\right)b. $$ In the right hand sum there are all the conjugates of $a$ apart from $a$ itself. Therefore $$ \sum_{i=1}^k(a^{2^i}+a^{2^{m-i}})=\sum_{i=0}^{m-1}a^{2^i}-a=tr(a)-a=tr(a)+a, $$ and the bilinear form is $$ B(a,b)=tr(tr(a)b+ab)=tr(ab)+tr(a)tr(b). $$ We are to prove that this bilinear form has maximal rank. Because we are in characteristic two, the form $B(a,b)$ is also symplectic. It is known that the rank of a symplectic form is always even. Here our space has odd dimension, so the radical of the form $$ R=\{b\in\mathbb{F}_{2^m}\mid B(a,b)=0\ \text{for all $a\in \mathbb{F}_{2^m}$}\} $$ must be at least one-dimensional. Indeed, we observe that $b=1$ is in the radical, as $$ B(a,1)=tr(a\cdot1)+tr(a)tr(1)=tr(a)+tr(a)\cdot m=tr(a)(1+2k+1)=0. $$ The remaining task is thus to show that $R=\{0,1\}$. So let $b\in\mathbb{F}_{2^m}$, $b\neq0,1$. Consider the bilinear form as a polynomial function of the first variable $x$ $$ B(x,b)=tr(xb)+tr(x)tr(b)=\sum_{i=0}^{m-1}b^{2^i}x^{2^i}+\sum_{i=0}^{m-1}tr(b)x^{2^i} =\sum_{i=0}^{m-1}(tr(b)+b^{2^i})x^{2^i}. $$ Because $b\neq0,1$ we have here also $b^{2^{m-1}}\neq0,1$. Therefore $tr(b)+b^{2^{m-1}}\neq0$ and $B(x,b)$ is a polynomial of degree $2^{m-1}$ in the unknown $x$. Therefore $B(x,b)=0$ for at most $2^{m-1}$ values of $x$. As $a$ ranges over all of $\mathbb{F}_{2^m}$ it follows that $B(a,b)\neq0$ for some $a$. Q.E.D.
