奇正整数下二项式系数求和恒等式的证明请求

Hey there! Let's break down how to prove this neat identity for odd positive integers ( n ):

$$\sum_{k\ \text{odd}}^{n}\binom{2n+1}{2k}=\begin{cases} \binom{2^n+1}{2}, & \text{if}\ n\ \text{mod}\ 4 =1\ \binom{2^n}{2}, & \text{if}\ n\ \text{mod}\ 4 =3 \end{cases}$$

Key Observations & Setup #

First, note that when ( k ) is odd, ( 2k = 4s + 2 ) for some integer ( s \geq 0 ). So we're actually summing binomial coefficients of ( \binom{2n+1}{t} ) where ( t \equiv 2 \pmod{4} ). We'll use complex roots of unity (specifically the imaginary unit ( i )) and properties of the binomial theorem to isolate these terms.

We start with two key facts:

• The sum of all even-indexed binomial coefficients for ( \binom{2n+1}{t} ) is ( 2^{2n} ) (since the total sum of all binomial coefficients is ( 2^{2n+1} ), splitting evenly between even and odd indices).
• For any integer ( m ), ( (1+i)^m + (1-i)^m = \sum_{t=0}^m \binom{m}{t}(i^t + (-i)^t) ), which only retains terms where ( t ) is even (odd terms cancel out). For even ( t ):
  • If ( t = 4s ), ( i^t + (-i)^t = 2 )
  • If ( t = 4s+2 ), ( i^t + (-i)^t = -2 )

Step 1: Express the Target Sum in Terms of Complex Powers #

Let ( S = \sum_{k\ \text{odd}}^{n}\binom{2n+1}{2k} ) (our target sum). The sum of even-indexed coefficients with ( t \equiv 0 \pmod{4} ) is ( 2^{2n} - S ). Plugging into the complex power identity:
$$(1+i)^{2n+1} + (1-i)^{2n+1} = 2\left[(2^{2n} - S) - S\right] = 4(2^{2n-1} - S)$$
Rearranging to solve for ( S ):
$$S = 2^{2n-1} - \frac{1}{4}\left[(1+i)^{2n+1} + (1-i)^{2n+1}\right]$$

Step 2: Simplify the Complex Terms #

We can rewrite ( (1+i)^{2n+1} ) and ( (1-i)^{2n+1} ) using ( (1+i)^2 = 2i ) and ( (1-i)^2 = -2i ):
$$(1+i)^{2n+1} = (1+i)(2i)^n = 2^n(1+i)in$$
$$(1-i)^{2n+1} = (1-i)(-2i)^n = 2^n(1-i)(-i)n$$
Combining these:
$$(1+i)^{2n+1} + (1-i)^{2n+1} = 2^{n\left[(1+i)i}n + (1-i)(-i)^n\right]$$

Step 3: Case Analysis Based on ( n \mod 4 ) #

Since ( n ) is odd, we only need to consider two cases:

Case 1: ( n \equiv 1 \pmod{4} )

Here, ( i^n = i ) and ( (-i)^n = -i ). Substitute into the expression:
$$(1+i)i + (1-i)(-i) = i + i^2 - i + i^2 = -1 -1 = -2$$
So:
$$(1+i)^{2n+1} + (1-i)^{2n+1} = 2^n(-2) = -2^{n+1}$$
Plug back into ( S ):
$$S = 2^{2n-1} - \frac{1}{4}(-2^{n+1}) = 2^{2n-1} + 2^{n-1} = \frac{2ⁿ⁽²n + 1)}{2} = \binom{2^n + 1}{2}$$

Case 2: ( n \equiv 3 \pmod{4} )

Here, ( i^n = -i ) and ( (-i)^n = i ). Substitute into the expression:
$$(1+i)(-i) + (1-i)i = -i -i^2 + i -i^2 = 1 + 1 = 2$$
So:
$$(1+i)^{2n+1} + (1-i)^{2n+1} = 2^n(2) = 2^{n+1}$$
Plug back into ( S ):
$$S = 2^{2n-1} - \frac{1}{4}(2^{n+1}) = 2^{2n-1} - 2^{n-1} = \frac{2ⁿ⁽²n - 1)}{2} = \binom{2^n}{2}$$

Conclusion #

Both cases match the identity we needed to prove!

内容的提问来源于stack exchange，提问作者Blind Miner