= Solution
Over $N$ <half-lives>, the duration is $T=Nt_{1/2}=N\ln2/\lambda$. The time-averaged thermal <power> is
$$
\overline P=\frac1T\int_0^T P_0e^{-\lambda t}\,dt
=\frac{P_0}{\lambda T}(1-e^{-\lambda T}).
$$
Since $\lambda T=N\ln2$ and $e^{-\lambda T}=2^{-N}$,
$$
\boxed{\overline P=\frac{P_0}{N\ln2}(1-2^{-N}).}
$$
This is a time average, not the arithmetic mean of the initial and final powers. It follows directly from the assumed constant <nuclear decay constant> and constant recovered energy per decay.
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