Problems

Solution:

1. Notice that \(1 \cdot 3 \cdot 5 \cdot 7 \cdot (2n-1) = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots \cdots 2n}{2 \cdot 4 \cdot 6 \cdots 2n} = \frac{(2n)!}{2^n \cdot n!}\) as required. When \(|4x| < 1\) or \(|x|<\frac14\) we can apply the generalised binomial theorem to see that: \begin{align*} \frac{1}{\sqrt{1-4x}} &= (1-4x)^{-\frac12} \\ &= 1+\sum_{n=1}^\infty \frac{-\frac12 \cdot \left ( -\frac32\right)\cdots \left ( -\frac{2n-1}2\right)}{n!} (-4x)^n \\ &= 1+\sum_{n=1}^{\infty} (-1)^n\frac{(2n)!}{(n!)^2 2^{2n}} (-4)^n x^n \\ &= 1+\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2 } x^n \\ \end{align*}
2. Differentiating we obtain \begin{align*} && 2(1-4x)^{-\frac32} &= \sum_{n=1}^\infty \frac{(2n)!}{n!(n-1)!} x^{n-1} \\ \Rightarrow &&\sum_{n=1}^\infty \frac{(2n)!}{n!(n-1)!} \left (\frac{6}{25} \right)^{n} &= \frac{6}{25} \cdot 2\left(1- 4 \frac{6}{25}\right)^{-\frac32} \\ &&&= \frac{12}{25} \left (\frac{1}{25} \right)^{-\frac32} \\ &&&= \frac{12}{25} \cdot 125 = 60 \end{align*}
3. By integrating, we obtain \begin{align*} && \int_{t=0}^{t=x} \frac{1}{\sqrt{1-4t}} \d t &= \int_{t=0}^{t=x} \left (1+\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2 } t^n \right) \d t \\ \Rightarrow && \left [ -\frac12 \sqrt{1-4t}\right]_0^x &= x + \sum_{n=1}^{\infty} \frac{(2n)!}{n!(n+1)! } x^{n+1} \\ \Rightarrow && \frac12 - \frac12\sqrt{1-4x} - x &= \sum_{n=1}^{\infty} \frac{(2n)!}{n!(n+1)! } x^{n+1} \\ \\ \Rightarrow && 9\sum_{n=1}^{\infty} \frac{(2n)!}{n!(n+1)! } \left ( \frac{2}{9}\right)^{n+1} &=9 \cdot\left( \frac12 - \frac12 \sqrt{1-4\cdot \frac{2}{9}} - \frac29\right )\\ &&&= 9 \cdot \left (\frac12 - \frac{1}{6} - \frac{2}{9} \right) \\ &&&= 1 \end{align*}