# 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 $b$ ?

$$(a+b)^n = \sum_{k=0}^n {n\choose k} b^k a^{n-k} \quad\text{and}\quad(a-b)^n = \sum_{k=0}^n {n\choose k} (-1)^{k}b^k a^{n-k}$$ So that we have, $$(b+a)^n-(b-a)^n = \sum_{k=0}^n {n\choose k}\color{red}{ \left[1-(-1)^{k}\right]}a^k b^{n-k} =\color{red}{2}\sum_{k \text{ odd}} {n\choose k} a^k b^{n-k}$$

• Ah! Silly me. Thanks! I came across another way of obtaining a closed form expression for a particular case when $a = p$ and $b = 1-p$ that involves SVD on a matrix. I was just wondering if a more straightforward approach is there! Commented Nov 20, 2017 at 10:41
• @ShreyasR see the edit what happen if you take only odd k? in the red braket? Commented Nov 20, 2017 at 10:45
• Oh, since when it is even, you get zero? Commented Nov 20, 2017 at 10:47
• @AirConditioner of course. Commented Nov 20, 2017 at 10:48
• $\quad(a-b)^n = \sum_{k=0}^n {n\choose k} (-1)^{k}b^k a^{n-k}$ I think this should be $\sum_{k=0}^n {n\choose k} (-b)^{k} a^{n-k}$ as the other one does not expand properly in mathematica Commented Oct 30, 2022 at 13:32

Hint: expand $(a+b)^n$ and $(a-b)^n$. What happens when you take the sum (or the difference) of the two?

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.