= Solution
A <K correction> accounts for observing a redshifted spectrum through a fixed <photometric passband>, rather than the corresponding emitted band. In the same R-band <astronomical magnitude> convention,
$$
M_R=m_R-5\log_{10}(d_L/10\,\mathrm{pc})-K_R.
$$
The <luminosity distance> already incorporates bolometric dimming; the additional correction depends on spectral shape and band response. At $z=3$, observed R-band light at $660\,\mathrm{nm}$ was emitted at $165\,\mathrm{nm}$, in the far ultraviolet. It is not a direct measurement of rest-frame R-band luminosity.
To fix the sign, let $L_\nu$ be emitted spectral <luminosity> and approximate R as narrow. Counting photon energy, arrival time and the transformed frequency interval gives $F_{\nu_o}=(1+z)L_{(1+z)\nu_o}/(4\pi d_L^2)$. Thus
$$
K_R=-2.5\log_{10}\!\left[(1+z)
\frac{L_\nu((1+z)\nu_R)}{L_\nu(\nu_R)}\right],
\qquad M_{R,\rm ignored}-M_R=K_R.
$$
For an ordinary old <elliptical galaxy>, the far-ultraviolet spectrum is weak relative to the optical, so typically $L_\nu(165\,\mathrm{nm})/L_\nu(660\,\mathrm{nm})<1/4$. Then $K_R>0$: \b[neglecting it overestimates the numerical absolute magnitude and underestimates the R-band luminosity]. For a young, unobscured <star-forming galaxy>, hot stars make the far-ultraviolet strong, with an approximately flat $L_\nu$ spectrum often serving as a useful model. It gives $K_R=-2.5\log_{10}4\simeq-1.51$: \b[neglecting it underestimates the numerical absolute magnitude and overestimates the R-band luminosity]. Strong dust extinction or an unusual spectrum can reverse this latter sign; a galaxy label alone does not prove an exact correction. The band-integrated convention is given in https://arxiv.org/abs/astro-ph/0210394[Hogg and collaborators' definition of the K-correction].
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