Let $w_n$ denote an arbitrary word of $n$ elements. Let $a_n$ denote the number of ways $W_n$ elements may be parenthesized. First note that we have: {{begin-eqn}} {{eqn | l = a_1       | r = 1       | c =  }} {{eqn | l = a_2       | r = 1       | c =  }} {{eqn | l = a_3       | r = 2       | c = that is, $b_1 \paren {b_2 b_3}$ and $\paren {b_1 b_2} b_3$ }} {{end-eqn}} and from Parenthesization of Word of $4$ Elements: :$a_4 = 5$ Consider a word $w_{n + 1}$ of $n + 1$ elements. Then $w_{n + 1}$ can be formed as any one of: :$w_1$ concatenated with $w_n$ :$w_2$ concatenated with $w_{n - 1}$ :$\dotsc$ and so on until: :$w_n$ concatenated with $w_1$ Thus the $i$th row in the above sequence is the number of parenthesizations of $w_{n + 1}$ in which the two outermost parenthesizations contain $i$ and $n - i + 1$ terms respectively. We have that: :there are $a_i$ parenthesizations of $w_i$ :there are $a_{n - i + 1}$ parenthesizations of $w_{n - i + 1}$ Hence the total number of parenthesizations of $w_{n + 1}$ is the sum of all these parenthesizations for $1 \le i \le n$. That is: :$(1): \quad a_{n + 1} = a_1 a_n + a_2 a_{n - 1} + \dotsb + a_n a_1$ Let us start with the generating function: :$\map {G_A} z = \displaystyle \sum_{n \mathop = 1}^\infty a_n z^n$ Then: {{begin-eqn}} {{eqn | l = \map {G_A} z       | r = z + \sum_{n \mathop = 2}^\infty \paren {a_1 a_n + a_2 a_{n - 1} + \dotsb + a_n a_1} z^n       | c = from $(1)$ }} {{eqn | r = z + \sum_{n \mathop = 1}^\infty a_n z^n \sum_{n \mathop = 1}^\infty a_n z^n       | c =  }} {{eqn | r = z + \paren {\map {G_A} z}^2       | c =  }} {{end-eqn}} Thus $\map {G_A} z$ satisfies the quadratic equation: :$\paren {\map {G_A} z}^2 - \map {G_A} z + z = 0$ By the Quadratic Formula, this gives: :$\map {G_A} z = \dfrac {1 \pm \sqrt {1 - 4 z} } 2$ Since $\map {G_A} 0 = 0$, we can eliminate the positive square root and arrive at: :$(2): \quad \map {G_A} z = \dfrac 1 2 - \dfrac {\sqrt {1 - 4 z} } 2$ Expanding $\sqrt {1 - 4 z}$ using the Binomial Theorem: :$\map {G_A} z = \dfrac 1 2 - \dfrac 1 2 \displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \dbinom {\frac 1 2} n 4^n z^n$ where: :$\dbinom {\frac 1 2} 0 = 1$ and: :$\dbinom {\frac 1 2} n = \dfrac {\frac 1 2 \paren {\frac 1 2 - 1} \dotsm \paren {\frac 1 2 - n + 1} } {n!}$ As a result: :$(3): \quad \map {G_A} z = -\dfrac 1 2 \displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^n \dbinom {\frac 1 2} n 4^n z^n$ We can expand $(3)$ as a Taylor series about $0$. As such a series, when it exists, is unique, the coefficients must be $a_n$. Hence: {{begin-eqn}} {{eqn | l = a_n       | r = -\dfrac 1 2 \paren {-1}^n \dbinom {\frac 1 2} n 4^n       | c =  }} {{eqn | r = \paren {-1}^{n - 1} \dfrac 1 2 \dfrac {\frac 1 2 \paren {\frac 1 2 - 1} \dotsm \paren {\frac 1 2 - n + 1} } {n!} 4^n       | c =  }} {{eqn | r = \paren {-1}^n \dfrac 1 2 \dfrac {\paren {-1} \paren {1 - 2} \dotsm \paren {1 - 2 \paren {n - 1 } } 2^n} {n!}       | c =  }} {{eqn | r = \dfrac 1 2 \dfrac {1 \times 3 \times \dotsb \times \paren {2 n - 3} 2^n} {n!}       | c =  }} {{eqn | r = \dfrac 1 2 \dfrac {1 \times 3 \times \dotsb \times \paren {2 n - 3} n! 2^n} {n! n!}       | c = multiplying top and bottom by $n!$ }} {{eqn | r = \dfrac 1 2 \dfrac {1 \times 2 \times 3 \times 4 \times \dotsb \times \paren {2 n - 4} \paren {2 n - 3} \paren {2 n - 2} \paren {2 n} } {\paren {n!}^2}       | c =  }} {{eqn | r = \dfrac 1 2 \dfrac {1 \times 2 \times 3 \times 4 \times \dotsb \times \paren {2 n - 2} } {\paren {\paren {n - 1}!}^2}       | c =  }} {{eqn | r = \frac 1 n \binom {2 n - 2} {n - 1}       | c =  }} {{end-eqn}} {{qed}}