# Expansion with binomial coefficients

How can I get binomial coefficients in expansion of $(n x+i) (1+i x)^n+(n x-i) (1-i x)^n$, where $i=\sqrt{-1}$ and $n$ is an integer. I have no idea how to coax Mathematica to do something remotely close to this without doing a lot of copying and pasting as if I was doing it by hand. If possible, I like the coefficients given in binomial notation.

• Would help if actual Mathematica input was provided.. Jan 8 '14 at 17:58
• you mean: (1 - I x)^n (-I + n x) + (1 + I x)^n (I + n x)?
– user11356
Jan 8 '14 at 18:00

It is difficult to automatically obtain such expansion. However, we can make сhain of identical transformations

I (1 - I n x) (1 + I x)^n - I (1 + I n x) (1 - I x)^n // FullSimplify

I (1 - I n x) Sum[Binomial[n, m] (I x)^m, {m, 0, n}] -
I (1 + I n x) Sum[Binomial[n, m] (-I x)^m, {m, 0, n}] // FullSimplify

I (1 - I n x) Sum[Binomial[n, m] (I x)^m, {m, 0, n}] -
I (1 + I n x) Sum[Binomial[n, m] (-I x)^m, {m, 0, n}] // FullSimplify

2 I Sum[Binomial[n, m] (I x)^m, {m, 1, n, 2}] +
2 n x Sum[Binomial[n, m] (I x)^m, {m, 0, n, 2}] // FullSimplify

2 I Sum[Binomial[n, m] (I x)^m, {m, 1, n, 2}] -
2 n I Sum[Binomial[n, m - 1] (I x)^m, {m, 1, n + 1, 2}] // FullSimplify

Sum[2 I (Binomial[n, m] - n Binomial[n, m - 1]) (I x)^m, {m, 1, n + 1,
2}] // FullSimplify

Sum[2 I (Binomial[n, 2 j + 1] - n Binomial[n, 2 j]) (I x)^(2 j + 1),
{j, 1, (n + 1)/2}] // FullSimplify


Each line outputs

(1 - I x)^n (-I + n x) + (1 + I x)^n (I + n x)


Finally, the expansion is

Sum[2 I (Binomial[n, 2 j + 1] - n Binomial[n, 2 j]) (I x)^(2 j + 1),
{j, 1, (n + 1)/2}] // HoldForm // TraditionalForm


$$\sum _{j=1}^{\frac{n+1}{2}} 2 i \left(\binom{n}{2 j+1}-n \binom{n}{2 j}\right) (i x)^{2 j+1}$$

• That's very nice. Thank you. It is easy to see that those pesky $i$'s go away. I did get the same expression doing it by hand (of course with no $i$'s in it). I wish I could also accept this as the answer.
– user11356
Jan 8 '14 at 23:28
• With all due respect to Daniel, I think this reply meets the conditions I set in my question better. So, I'll switch my accepted answer. I wish I could accept both.
– user11356
Jan 9 '14 at 0:32
• Heartbroken though I am over the loss of credit, I'm fully in agreement with the outcome here. (Actually I thought of noting yesterday that you might have accepted my response too soon, and should feel free to undo that. I would have sent it today had you not made the change.) Jan 9 '14 at 16:29

If you mean to do this for general n then this is something that "works", after a fashion.

ee = (1 - I x)^n (-I + n x) + (1 + I x)^n (I + n x);

ss =
SeriesCoefficient[ee, {x, 0, j},
Assumptions -> 0 <= j <= n && Element[n, Integers]];


The result is in terms of ghastly DifferenceRoot objects. It evaluates nicely though, and in a way that gives functions of n (as it ought).

Table[ss, {j, 0, 10}]

(* Out= {0, 0, 0, -2 (-(1/6) (-2 + n) (-1 + n) n +
1/2 (-1 + n) n^2), 0,
2 (-(1/120) (-4 + n) (-3 + n) (-2 + n) (-1 + n) n +
1/24 (-3 + n) (-2 + n) (-1 +
n) n^2), 0, -2 (-(((-6 + n) (-5 + n) (-4 + n) (-3 + n) (-2 +
n) (-1 + n) n)/5040) +
1/720 (-5 + n) (-4 + n) (-3 + n) (-2 + n) (-1 + n) n^2), 0,
2 (-(1/362880)(-8 + n) (-7 + n) (-6 + n) (-5 + n) (-4 + n) (-3 +
n) (-2 + n) (-1 + n) n + ((-7 + n) (-6 + n) (-5 + n) (-4 +
n) (-3 + n) (-2 + n) (-1 + n) n^2)/40320), 0} *)

• Thank you. That's a lot more than what I was able to accomplish. I need to study your reply tho :)
– user11356
Jan 8 '14 at 18:16
• Now that I understand what that output means, it seems I have to it by hand. But I thank you for taking the time to answer.
– user11356
Jan 8 '14 at 18:35
• Unless you are versed in DifferenceRoot understanding, I'd not give that aspect much thought (I'm not, and I didn't). Just treat it as a black box is what I recommend (and do). Jan 8 '14 at 18:41