How do you drop to 50% power with just one dead cylinder? ISA conditions. Ignore the prop for now.
As another poster pointed out, a bunch of energy is wasted in pumping that dead piston. It's not free energy doing that.
From Aircraft Systems for Pilots," by Dale DeRemer, PH.D., the college-level textbook for aircraft systems courses, gives us figures on thermal efficiency for the aircraft reciprocating engine:
Of the total heat produced, 15 to 20% is removed by cooling (heat radiated from cylinder head fins and oil cooler); 5 to 10% is utilized in overcoming friction of moving parts, most of which produces heat that is carried away by the lubricating oil; and 40 to 45% is lost through the exhaust.
The remaining heat energy drives the engine, and it's not great, typically around 25 to 35%. We lose a quarter of that with a four-banger's dead cylinder, and expend some of the rest in pushing the dead stuff back and forth and in pumping air in and out of it.
When I was instructing we had to demonstrate a partial power loss after takeoff. The drill was to reduce the throttle to get a 100-RPM drop, and that was enough, in a 172, to cut the climb rate by about half. It's serious stuff. And it's due to the nonlinear loss of thrust when you lose RPM. The propeller is at its most efficient when it is near redline RPM, and we can see that if we're on the ground and throttle up to half of redline; the airplane will taxi fast but it will never lift off. At half of that we're at idle and the airplane won't even move much at all.
When we lose even that much power (100 RPM), we have to either drop the nose to maintain Vy, such as it may be. If we keep the nose up we increase the AoA because the power isn't there to keep the airspeed at that attitude, and drag goes up. If we have a constant-speed prop we might get the same RPM, but at a lower pitch setting, which reduces airspeed and you still have to drop the nose. You lose climb rate anyway.