Problem source

Given that $y=\cosh(n\cosh^{-1}x),$ for $x\geqslant1,$ prove that
\[
y=\frac{(x+\sqrt{x^{2}-1})^{n}+(x-\sqrt{x^{2}-1})^{n}}{2}.
\]
Explain why, when $n=2k+1$ and $k\in\mathbb{Z}^{+},$ $y$ can also be expressed as the polynomial 
\[
a_{0}x+a_{1}x^{3}+a_{2}x^{5}+\cdots+a_{k}x^{2k+1}.
\]
Find $a_{0},$ and show that 
\begin{questionparts}
\item $a_{1}=(-1)^{k-1}2k(k+1)(2k+1)/3$; 
\item $a_{2}=(-1)^{k}2(k-1)k(k+2)(2k+1)/15.$
\end{questionparts}
Find also the value of ${\displaystyle \sum_{r=0}^{k}a_{r}.}$

Solution source

Recall, $\cosh^{-1} x = \ln (x + \sqrt{x^2-1})$

\begin{align*}
\cosh(n \cosh^{-1} x) &= \frac12 \left ( \exp(n \cosh^{-1} x) + \exp(-n\cosh^{-1}x) \right) \\
&= \frac12 \left ((x + \sqrt{x^2-1})^n + (x + \sqrt{x^2-1})^{-n} \right) \\
&= \frac12 \left ((x + \sqrt{x^2-1})^n + (x - \sqrt{x^2-1})^{n} \right) \\
\end{align*}

When $n = 2k+1$

\begin{align*}
\cosh(n \cosh^{-1} x)&=  \frac12 \left ((x + \sqrt{x^2-1})^n + (x - \sqrt{x^2-1})^{n} \right) \\
&= \frac12 \left (\sum_{i=0}^{2k+1}\binom{2k+1}{i}x^{2k+1-i}\left ( (\sqrt{x^2-1}^{i} + (-\sqrt{x^2-1})^{i}   \right)  \right) \\
&=\sum_{i=0}^{k} \binom{2k+1}{2i}x^{2k+1-2i}(x^2-1)^i \\
&=\sum_{i=0}^{k} \binom{2k+1}{2i}x^{2(k-i)+1}(x^2-1)^i \\
\end{align*}

Which is clearly a polynomial with only odd degree terms.

\begin{align*}
a_0 &= \frac{\d y}{\d x} \vert_{x=0} \\
&= \sum_{i=0}^k\binom{2k+1}{2i} \left ( (2(k-i)+1)x^{2(k-i)}(x^2-1)^i + 2i\cdot x^{2(k-i)+2}(x^2-1) \right) \\
&= \binom{2k+1}{2k} (-1)^{k} \\
&= (-1)^k(2k+1)
\end{align*}

\begin{questionparts}
\item \begin{align*}
a_1 &= \binom{2k+1}{2k}\binom{k}{1}(-1)^{k-1}+\binom{2k+1}{2(k-1)}(-1)^{k-1} \\
&=(-1)^{k-1}\cdot ( (2k+1)k + \frac{(2k+1)\cdot 2k \cdot (2k-1)}{3!}) \\
&= (-1)^{k-1}(2k+1)k\frac{3 + 2k-1}{3} \\
&= (-1)^{k-1}2(2k+1)k (k+1) 
\end{align*}
\item \begin{align*}
a_2 &= \binom{2k+1}{2k} \binom{k}{2}(-1)^{k-2} + \binom{2k+1}{2(k-1)} \binom{k-1}{1} (-1)^{k-2}+\binom{2k+1}{2(k-2)} (-1)^{k-2} \\
&= \binom{2k+1}{1} \binom{k}{2}(-1)^{k-2} + \binom{2k+1}{3} \binom{k-1}{1} (-1)^{k-2}+\binom{2k+1}{5} (-1)^{k-2} \\
&= (-1)^{k} \left (\binom{2k+1}{1} \frac{k(k-1)}{2} + \binom{2k+1}{3}(k-1)+\binom{2k+1}{5}  \right) \\
&= (-1)^{k} \left ( \frac{(2k+1)k(k-1)}{2} + \frac{(2k+1)k(2k-1)}{3} + \frac{(2k+1)k(2k-1)(k-1)(2k-3)}{5\cdot2\cdot3} \right) \\
&= (-1)^k (2k+1)k\frac{1}{30} \left ( 15(k-1) + 10(2k-1)+(2k-1)(k-1)(2k-3) \right)
\end{align*}
\end{questionparts}

\begin{align*}
\sum_{r=0}^k a_k &= \frac12 \left ((1 + \sqrt{1^2-1})^n + (1 - \sqrt{1^2-1})^{n} \right) \\
&= 1
\end{align*}