= Solution
The product-to-sum identity gives $V_1=V_0$, $V_2=V_0/2$ and $V_3=2V_0/3$, with all other modal coefficients zero. If $\mu$ is the string's <linear mass density>, the initial <potential energy> vanishes and <orthogonality> gives the <modal energy of a string>
$$
E_n=\frac\mu2\int_0^LV_n^2\sin^2\frac{n\pi x}{L}\,dx=\frac{\mu L}{4}V_n^2.
$$
Consequently $\boxed{E_1:E_2:E_3=36:9:16}$ for $V_0\ne0$. \b[The fundamental mode carries the most energy], with fraction $36/61$. Comparing displacement coefficients without their <frequency> factors would give the wrong energy test.
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