 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 stringConsequently $\boxed{E_1:E_2:E_3=36:9:16}$ for $V_0\ne 0$. The fundamental mode carries the most energy, with fraction $36/61$. Comparing displacement coefficients without their frequency factors would give the wrong energy test.