Using Equation 3.114,
\left\langle n|x| n^{\prime}\right\rangle=\sqrt{\frac{\hbar}{2 m \omega}}\left(\sqrt{n^{\prime} \delta}_{n, n^{\prime}-1}+\sqrt{n} \delta_{n^{\prime}, n-1}\right) (3.114).
\langle x\rangle=\int\left(\sum_{n} c_{n} \psi_{n} e^{-i E_{n} t / \hbar}\right)^{*} x\left(\sum_{n^{\prime}} c_{n^{\prime}} \psi_{n^{\prime}} e^{-i E_{n^{\prime}} t / \hbar}\right) d x=\sum_{n} \sum_{n^{\prime}} c_{n}^{*} c_{n^{\prime}} e^{i\left(E_{n}-E_{n^{\prime}}\right) t / \hbar}\left\langle n|x| n^{\prime}\right\rangle .
=\sqrt{\frac{\hbar}{2 m \omega}} \sum_{n} \sum_{n^{\prime}} c_{n}^{*} c_{n^{\prime}} e^{i\left(E_{n}-E_{n^{\prime}}\right) t / \hbar}\left(\sqrt{n^{\prime}} \delta_{n, n^{\prime}-1}+\sqrt{n} \delta_{n^{\prime}, n-1}\right) .
=\sqrt{\frac{\hbar}{2 m \omega}} \sum_{n}\left[\sqrt{n+1} c_{n}^{*} c_{n+1} e^{i\left(E_{n}-E_{n+1}\right) t / \hbar}+\sqrt{n} c_{n}^{*} c_{n-1} e^{i\left(E_{n}-E_{n-1}\right) t / \hbar}\right] .
=\sqrt{\frac{\hbar}{2 m \omega}}\left[\sum_{n=0}^{\infty} \sqrt{n+1} c_{n}^{*} c_{n+1} e^{-i \omega t}+\sum_{n=0}^{\infty} \sqrt{n} c_{n}^{*} c_{n-1} e^{i \omega t}\right] .
In the last step I used
E_{n}-E_{n-1}=\left(n+\frac{1}{2}\right) \hbar \omega-\left(n-\frac{1}{2}\right) \hbar \omega=\hbar \omega, \quad E_{n}-E_{n+1}=\left(n+\frac{1}{2}\right) \hbar \omega-\left(n+\frac{3}{2}\right) \hbar \omega=-\hbar \omega .
Noting that \sqrt{n} kills the n = 0 term in the second sum, let n → n + 1 and run it from 0 to ∞:
\langle x\rangle=\sqrt{\frac{\hbar}{2 m \omega}} \sum_{n=0}^{\infty} \sqrt{n+1}\left[c_{n}^{*} c_{n+1} e^{-i \omega t}+c_{n+1}^{*} c_{n} e^{i \omega t}\right] .
=\left(\sqrt{\frac{\hbar}{2 m \omega}} \sum_{n=0}^{\infty} \sqrt{n+1} c_{n}^{*} c_{n+1}\right) e^{-i \omega t}+\left(\sqrt{\frac{\hbar}{2 m \omega}} \sum_{n=0}^{\infty} \sqrt{n+1} c_{n} c_{n+1}^{*}\right) e^{i \omega t} .
=\frac{1}{2}\left(C e^{i \phi}\right) e^{-i \omega t}+\frac{1}{2}\left(C e^{-i \phi}\right) e^{i \omega t}=C \cos (\omega t-\phi) .
In the case of Problem 2.40, c_{0}=3 / 5, c_{1}=-2 \sqrt{2} / 5, c_{2}=2 \sqrt{2} / 5 , (and all the rest are zero) so
C e^{-i \phi}=\sqrt{\frac{2 \hbar}{m \omega}}\left(c_{1}^{*} c_{0}+\sqrt{2} c_{2}^{*} c_{1}\right)=\sqrt{\frac{2 \hbar}{m \omega}}\left(\frac{(-2 \sqrt{2})}{5} \frac{3}{5}+\sqrt{2} \frac{2 \sqrt{2}}{5} \frac{(-2 \sqrt{2})}{5}\right)=-\frac{28}{25} \sqrt{\frac{\hbar}{m \omega}} .
C=\frac{28}{25} \sqrt{\frac{\hbar}{m \omega}}, \quad \phi=\pi .