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Theorem
$\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 0}^{n} \binom {2n} {2k} E_{2n - 2k } = 0$
where $E_k$ denotes the $k$th Euler number.
Proof
Take the definition of Euler numbers:
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\(\ds \sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!}\)
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\(=\)
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\(\ds \frac {2e^x} {e^{2x} + 1}\)
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\(\ds \)
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\(=\)
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\(\ds \paren {\frac {2e^x } {e^{2x} + 1 } } \paren {\frac {e^{-x } } {e^{-x } } }\)
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Multiply by 1
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\(\ds \)
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\(=\)
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\(\ds \paren {\frac 2 {e^{x} + e^{-x} } }\)
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From the definition of the exponential function:
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\(\ds e^x\)
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\(=\)
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\(\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\)
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\(\ds \)
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\(=\)
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\(\ds 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \frac {x^4} {4!} + \cdots\)
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\(\ds e^{-x}\)
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\(=\)
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\(\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-x }^n } {n!}\)
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\(\ds \)
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\(=\)
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\(\ds 1 - x + \frac {x^2} {2!} - \frac {x^3} {3!} + \frac {x^4} {4!} - \cdots\)
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\(\ds \paren {\frac {e^x + e^{-x } } 2 }\)
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\(=\)
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\(\ds \paren {\sum_{n \mathop = 0}^\infty \frac {x^{2n } } {\paren {2n }!} }\)
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\(\ds \)
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\(=\)
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\(\ds 1 + \frac {x^2} {2!} + \frac {x^4} {4!} + \cdots\)
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odd terms cancel in the sum.
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Thus:
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\(\ds 1\)
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\(=\)
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\(\ds \paren {\frac 2 {e^x + e^{-x } } } \paren {\frac {e^x + e^{-x } } 2 }\)
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\(\ds \)
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\(=\)
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\(\ds \paren {\sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {x^{2n } } {\paren {2n }!} }\)
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By Product of Absolutely Convergent Series, we will let:
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\(\ds a_n\)
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\(=\)
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\(\ds \frac {E_n x^n} {n!}\)
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\(\ds b_n\)
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\(=\)
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\(\ds \frac {x^{2n} } {\paren {2n }!}\)
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Then:
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\(\ds \sum_{n \mathop = 0}^\infty c_n\)
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\(=\)
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\(\ds \paren { \displaystyle \sum_{n \mathop = 0}^\infty a_n } \paren {\displaystyle \sum_{n \mathop = 0}^\infty b_n }\)
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\(\ds =1\)
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\(\ds c_n\)
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\(=\)
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\(\ds \sum_{k \mathop = 0}^n a_k b_{n - k}\)
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\(\ds c_0\)
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\(=\)
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\(\ds \frac {E_0 x^0} {0!} \frac {x^{0} } {0!}\)
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\(\ds = 1\)
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$c_0 = \paren {a_0 } \paren {b_{0 - 0 } } = \paren {a_0 } \paren {b_0 }$
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\(\ds \leadsto \ \ \)
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\(\ds \sum_{n \mathop = 1}^\infty c_n\)
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\(=\)
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\(\ds \paren { \displaystyle \sum_{n \mathop = 0}^\infty a_n } \paren {\displaystyle \sum_{n \mathop = 0}^\infty b_n } - a_0 b_0\)
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\(\ds =0\)
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Subtract 1 from both sides of the equation.
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$\forall n \in \Z_{\gt 0}$, term by term $c_n$ is equal to:
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\(\ds c_1\)
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\(=\)
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\(\ds \frac {E_0 x^0} {0!} \frac {x^2 } {2!} + \frac {E_1 x^1} {1!} \frac {x^0 } {0!}\)
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\(\ds = \frac {x^2 } {2! } E_0\)
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$= a_0 b_1 + a_1 b_0$
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\(\ds c_2\)
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\(=\)
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\(\ds \frac {E_0 x^0} {0!} \frac {x^4 } {4!} + \frac {E_1 x^1} {1!} \frac {x^2 } {2!} + \frac {E_2 x^2} {2!} \frac {x^0 } {0!}\)
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\(\ds = \frac {x^4 } {4! } E_0 + \frac {x^2 } {2! } E_2\)
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$= a_0 b_2 + a_1 b_1 + a_2 b_0$
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\(\ds c_3\)
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\(=\)
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\(\ds \frac {E_0 x^0} {0!} \frac {x^6 } {6!} + \frac {E_1 x^1} {1!} \frac {x^4 } {4!} + \frac {E_2 x^2} {2!} \frac {x^2 } {2!} + \frac {E_3 x^3} {3!} \frac {x^0 } {0!}\)
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\(\ds = \frac {x^6 } {6! } E_0 + \frac {x^4 } {2! 2! } E_2\)
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$= a_0 b_3 + a_1 b_2 + a_2 b_1 + a_3 b_0$
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\(\ds c_4\)
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\(=\)
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\(\ds \frac {E_0 x^0} {0!} \frac {x^8 } {8!} + \frac {E_1 x^1} {1!} \frac {x^6 } {6!} + \frac {E_2 x^2} {2!} \frac {x^4 } {4!} + \frac {E_3 x^3} {3!} \frac {x^2 } {2!} + \frac {E_4 x^4} {4!} \frac {x^0 } {0!}\)
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\(\ds = \frac {x^8 } {0! 8! } E_0 + \frac {x^6 } {2! 4! } E_2 + \frac {5 x^4 } {4! 0! } E_4\)
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$= a_0 b_4 + a_1 b_3 + a_2 b_2 + a_3 b_1 + a_4 b_0$
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\(\ds \cdots\)
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\(=\)
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\(\ds \cdots\)
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\(\ds c_n\)
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\(=\)
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\(\ds \frac {E_0 x^0} {0!} \frac {x^{2n} } {\paren {2n }!} + \frac {E_1 x^1} {1!} \frac {x^{2n - 2} } {\paren {2n - 2 }!} + \frac {E_2 x^2} {2!} \frac {x^{2n - 4} } {\paren {2n - 4 }!} + \cdots + \frac {E_n x^n} {n!} \frac {x^0 } {0!}\)
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Grouping like even terms produces:
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\(\ds \paren {\frac 1 {0! 2!} } E_0 + \paren {\frac 1 {2! 0!} } E_2\)
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\(=\)
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\(\ds 0\)
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$x^2$ term
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\(\ds \paren {\frac 1 {0! 4!} } E_0 + \paren {\frac 1 {2! 2!} } E_2 + \paren {\frac 1 {4! 0!} } E_4\)
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\(=\)
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\(\ds 0\)
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$x^4$ term
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Multiplying $c_n$ through by $n!$ gives:
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\(\ds n! c_n\)
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\(=\)
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\(\ds \frac {E_0 x^0} {0!} \frac {n! x^n } {n!} + \frac {E_1 x^1} {1!} \frac {n! x^{n-1} } {\paren {n - 1 }!} + \cdots + \frac {E_n x^n} {n!} \frac {n! x^{0} } {0!}\)
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\(\ds = 0\)
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\(\ds \)
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\(=\)
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\(\ds x^n \paren {\frac {n! } {0! n!} E_0 + \frac {n! } {1! \paren {n - 1 }!} E_1 + \cdots + \frac {n! } {n! 0!} E_n }\)
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\(\ds = 0\)
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factoring out $x^n$
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But those coefficients are the binomial coefficients:
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\(\ds n! c_n\)
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\(=\)
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\(\ds \dbinom n 0 E_0 + \dbinom n 1 E_1 + \dbinom n 2 E_2 + \cdots + \dbinom n n E_n\)
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\(\ds = 0\)
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Hence the result.
$\blacksquare$
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\(\ds \sum_{k \mathop = 0}^{n} \dbinom {2 n} {2 k} E_{2 n - 2 k}\)
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\(=\)
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\(\ds \binom {2 n} 0 E_{2 n} + \binom {2 n} 2 E_{2 n - 2} + \binom {2 n} 4 E_{2 n - 4} + \binom {2 n} 6 E_{2 n - 6} + \cdots + 1\)
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\(\ds = 0\)
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- $\begin{array}{r|cccccccccc}
B_k & \dbinom n 0 & & \dbinom n 1 & & \dbinom n 2 & & \dbinom n 3 & & \dbinom n 4 & & \dbinom n 5 \\
\hline
B_0 = 1 & 1 B_0 & & & & & & & & & & & = 1 \\
B_1 = -\frac 1 2 & 1 B_0 & + & 2 B_1 & & & & & & & & & = 0 \\
B_2 = +\frac 1 6 & 1 B_0 & + & 3 B_1 & + & 3 B_2 & & & & & & & = 0 \\
B_3 = 0 & 1 B_0 & + & 4 B_1 & + & 6 B_2 & + & 4 B_3 & & & & & = 0 \\
B_4 = -\frac 1 {30} & 1 B_0 & + & 5 B_1 & + & 10 B_2 & + & 10 B_3 & + & 5 B_4 & & & = 0 \\
B_5 = 0 & 1 B_0 & + & 6 B_1 & + & 15 B_2 & + & 20 B_3 & + & 15 B_4 & + & 6 B_5 & = 0 \\
\end{array}$
Also see