I need to develop $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i}$ using the binomial coefficients formula and show a few properties

Question

In the previous questions I've proved that $(1+i)^{2n+1}=a_n+ib_n$ where $a_n$ and $b_n$ are $\pm 2^n$ and that $(1-i)^{2n+1}=a_n - ib_n$ and that $|P_n(1)|=2^n$ using the previous statements. But now I need to show that the polynomial $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i}$ is of degree $2n$, even (the coefficient in front of $X^n$ is non-zero only when $n$ is even, with real coefficients, and with $2n+1$ as a dominant coefficient, using the binomial coefficient formula.

I also need to show that the coefficient $X^{2n-2}$ in $P_n$ is $\frac{n(1-2n)(1+2n)}{3}$.

I developed $(X+i)^{2n+1}$ and I get $X^{2n+1}+(2n+1)X^{2n}i-\frac{(2n+1)(2n)}{2}X^{2n-1}+...+(2n+1)X(-1)^n+(-1)^{n+1}$

When I develop $(X-i)^{2n+1}$ I get $X^{2n+1}+(2n+1)X^{2n}i-\frac{(2n+1)(2n)}{2}X^{2n-1}+...+(2n+1)X(-1)^n+(-1)^{n+1}$, so the same thing.

So obviously when I subtract the two parts I get $0$ which won't let me prove the properties. Where is my mistake ? Do I use the binomial coefficient the wrong way ?

EDIT: I think I have made a mistake in the second development, the second term should has a + sign when the subtraction is made so I get $\frac{2(2n+1)X^{2n}i}{2i} = (2n+1)X^{2n}$

How do I show it is even though ? Is the end result enough to show this property?

Answer 1

In the expansion $$(X+i)^{2n+1} - (X - i)^{2n+1}=\sum_{k=0}^{2n+1}\binom{2n+1}{k}X^ki^{2n+1-k}- \sum_{k=0}^{2n+1}\binom{2n+1}{k}X^k(-i)^{2n+1-k}$$ the terms such that $2n+1-k$ is even, i.e. k is odd, cancel pairwise, so there remains the terms with $k=2\ell+1\enspace (0\le \ell\le n)$, and the final expansion is \begin{align}(X+i)^{2n+1} &- (X - i)^{2n+1}=\sum_{\ell=0}^{n}\binom{2n+1}{2\ell} X^{2\ell}2\,i^{2(n-\ell)}\\ &=2\sum_{\ell=0}^{n}\binom{2n+1}{2\ell} X^{2\ell}(-1)^{n-\ell}. \end{align}

Answer 2

Since

$\begin{array}\\ (a+b)^n &=\sum_{k=0}^n \binom{n}{k}a^kb^{n-k}\\ (x+i)^{2n+1} &=\sum_{k=0}^{2n+1} \binom{2n+1}{k}x^ki^{2n+1-k}\\ &=\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}i^{2n+1-2k}+\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}i^{2n+1-(2k+1)}\\ &=\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}i^{2(n-k)+1}+\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}i^{2(n-k)}\\ &=i\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{n-k}+\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}(-1)^{n-k}\\ (x-i)^{2n+1} &=\sum_{k=0}^{2n+1} \binom{2n+1}{k}x^k(-1)^{2n+1-k}i^{2n+1-k}\\ &=\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{2n+1-2k}i^{2n+1-2k}+\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}(-1)^{2n+1-(2k+1)}i^{2n+1-(2k+1)}\\ &=\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{2(n-k)+1}i^{2(n-k)+1}+\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}(-1)^{2(n-k)}i^{2(n-k)}\\ &=-i\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{n-k}+\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}(-1)^{n-k}\\ \end{array} $

If we subtract the terms the second term cancels out and we are left with

$\begin{array}\\ (x+i)^{2n+1}-(x-i)^{2n+1} &=i\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{n-k}-(-i\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{n-k})\\ &=2i\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{n-k}\\ \text{so}\\ \dfrac{(x+i)^{2n+1}-(x-i)^{2n+1}}{2i} &=\sum_{k=0}^{n} \binom{2n+1}{2k}x^{2k}(-1)^{n-k}\\ \end{array} $

The highest order term has $k = n$ so it is $(2n+1)x^{2n} $.

If we add the terms instead of subtracting, the first term cancels out and we are left with

$\begin{array}\\ \dfrac{(x+i)^{2n+1}+(x-i)^{2n+1}}{2} &=\sum_{k=0}^{n} \binom{2n+1}{2k+1}x^{2k+1}(-1)^{n-k}\ \end{array} $

The highest order term has $k = n$ so it is $x^{2n+1} $.

Answer 3

From the binomial theorem, $$(x+a)^{2n+1}=\sum_{m=0}^{2n+1}{2n+1\choose m}x^ma^{2n+1-m}$$ so $$\begin{align} (x+a)^{2n+1}-(x-a)^{2n+1}&=\sum_{m=0}^{2n+1}{2n+1\choose m}x^m(a^{2n+1-m}-(-a)^{2n+1-m})\\ &=\sum_{m=0}^{2n+1}{2n+1\choose m}x^ma^{2n+1-m}(1-(-1)^{2n+1-m})\\ &=\sum_{m=0}^{2n+1}{2n+1\choose m}x^ma^{2n+1-m}(1+(-1)^{m}). \end{align}$$ Then, when $m$ is even, $(-1)^m=1$, and $1+(-1)^m=1+1=2$. Thus all the terms where $m=2k$ is even are non-zero, and we have $$\sum_{0\le m\le2n+1\\ m \text{ is even}}{2n+1\choose m}x^{m}a^{2n+1-m}(1+(-1)^{m})=\sum_{k=0}^{n}{2n+1\choose 2k}x^{2k}a^{2n+1-2k}(1+(-1)^{2k})\\ =2\sum_{k=0}^{n}{2n+1\choose 2k}x^{2k}a^{2n-2k+1}.$$ Then, when $m$ is odd, $(-1)^m=-1$, and $1+(-1)^m=1-1=0$. Thus all the terms where $m=2k+1$ is odd are $0$, and we have $$\sum_{0\le m\le2n+1\\ m \text{ is odd}}{2n+1\choose m}x^{m}a^{2n+1-m}(1+(-1)^{m})=\sum_{k=0}^{n}{2n+1\choose 2k+1}x^{2k+1}a^{2n+1-2k-1}(1+(-1)^{2k+1})\\ =0.$$ Then, $$\begin{align} (x+a)^{2n+1}-(x-a)^{2n-1}&=\sum_{0\le m\le2n+1\\ m \text{ is even}}{2n+1\choose m}x^{m}a^{2n+1-m}(1+(-1)^{m})\\ &+\sum_{0\le m\le2n+1\\ m \text{ is odd}}{2n+1\choose m}x^{m}a^{2n+1-m}(1+(-1)^{m})\\ &=\sum_{k=0}^{n}{2n+1\choose 2k}x^{2k}a^{2n+1-2k}(1+(-1)^{2k})\\ &=2\sum_{k=0}^{n}{2n+1\choose 2k}x^{2k}a^{2n-2k+1}. \end{align}$$ If you set $a=i$ you see that $a^{2n-2k+1}=(i^2)^{n-k}\cdot i=(-1)^{n-k}\cdot i$. Thus $$(x+i)^{2n+1}-(x-i)^{2n+1}=2i\sum_{k=0}^{n}(-1)^{n-k}{2n+1\choose 2k}x^{2k},$$ and, finally, $$p_n(x)=\frac1{2i}((x+i)^{2n+1}-(x-i)^{2n+1})=\sum_{k=0}^{n}(-1)^{n-k}{2n+1\choose 2k}x^{2k}.$$ You should be able to finish from here.