Actually this is not the way it works. Induced drag is very much a direct product of the amount of lift produced. The entire wing works together to produce the amount of lift the airplane needs at any moment. Since the total lift needed doesn't lessen or increase simply because the wing is twisted, that wing is going to be called on to generate whatever lift is required. And since the tips are not generating as much lift as they would without the twist, the lift has to be generated elsewhere. Since the wing generates lift by increasing the angle of attack, it simply has to increase more. So the inner parts of the wing wind up flying at a greater AOA than they could have had there been no washout. And they generate more induced drag than they would have at the lower AOA. The lift needed doesn't change simply because the wings are washed out. The induced drag that amount of lift generation creates doesn't change either.


This doesn't always work this way either. The energy of the tip vortex is very much a function of the lift the wing is being required to produce. That isn't going to change just because the tip is twisted or generating less of the overall lift required.


Very often us modelers "overdo" simple concepts. Very often the lift required at cruise needs very little AOA from the airfoil that's in the wing. The AOA might actually be less that the angle we've twisted into the poor wing. Since we've almost always placed our wing into the fuselage with an incidence that results in the wing flying at or very near it's cruise AOA then the entire wing is carried at cruise speed at the angle that would generate the appropriate lift with the least drag. But if we've twisted the tips a couple of degrees out of whack, what are they now doing? In a lot of cases that twisted tip is actually not lifting at all or is generating negative lift. In either case, the drag being created is added cost with no return.