Help with a bilinear form 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.
 A: 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.
