Sum of odd terms of a binomial expansion: $\sum\limits_{k \text{ odd}} {n\choose k} a^k b^{n-k}$ Is it possible to find a closed form expression for the sum
$$\sum_{k \text{ odd}} {n\choose k} a^k b^{n-k}$$
in terms of $a$ and $b$ ?
 A: $$(a+b)^n = \sum_{k=0}^n {n\choose k} b^k a^{n-k} \quad\text{and}\quad(a-b)^n = \sum_{k=0}^n {n\choose k} (-1)^{k}b^k a^{n-k} $$
So that we have,
$$(b+a)^n-(b-a)^n =  \sum_{k=0}^n {n\choose k}\color{red}{ \left[1-(-1)^{k}\right]}a^k b^{n-k} =\color{red}{2}\sum_{k \text{ odd}} {n\choose k} a^k b^{n-k}$$
A: Hint: expand $(a+b)^n$ and $(a-b)^n$. What happens when you take the sum (or the difference) of the two?
A: I found a rather interesting alternate method to solve this problem (from a problem in communication theory). Please share your thoughts.
Let 
$
  A = 
  \left[ {\begin{array}{cc}
   a & b \\
   b & a \\
  \end{array} } \right]
$
Now, $$[A^n]_{1,1} = \sum_{k \text{ even}} {n \choose k} a^k b^{n-k}$$
and $$[A^n]_{1,2} = \sum_{k \text{ odd}} {n \choose k} a^k b^{n-k}$$
$A$ can be decomposed as
$$  A = T^{-1}
  \left[ {\begin{array}{cc}
   a+b & 0 \\
   0 & a-b \\
  \end{array} } \right]T $$
where $ T = \left[ {\begin{array}{cc}
   1 & 1 \\
   1 & -1 \\
  \end{array} } \right] $
Now, $$  A^n = T^{-1}
  \left[ {\begin{array}{cc}
   (a+b)^n & 0 \\
   0 & (a-b)^n \\
  \end{array} } \right]T $$
$$  A^n = 
  \left[ {\begin{array}{cc}
   \frac{1}{2}[(a+b)^n + (a-b)^n] & \frac{1}{2}[(a+b)^n - (a-b)^n] \\
   \frac{1}{2}[(a+b)^n - (a-b)^n] & \frac{1}{2}[(a+b)^n + (a-b)^n] \\
  \end{array} } \right] $$
which gives the odd and even terms of the binomial expression.
