you would mow across one side, turn and mow next to your previous path, so that t=(nk)+n^2 thats the fastest way i see doing it.
This answer is partially correct! (But isn't totally) the question itself was rather difficult though.
The amount of corners you would take would be (2n - 2) not n, (if you had a 2x2 field you would have to cross two corners (As turning 180 degrees counts as one corner), if you had a three by three field you would need to cross 4 corners). Second, the amount of lines wouldn't be n
2 because you don't count the corners as a straight edge, so the answer would actually be n(n - 2) + 2. So the final equation for a line by line method would actually be...
(2n - 2)k + (n(n - 2) + 2)t, or
Equation 1) 2k(n - 1) + t(n(n - 2) + 2) (If you don't believe me try it out for n = 1, 2, 3, 4, etc)
The other thing is that isn't actually the ONLY fastest method, if we assume you don't have to turn for the last corner (we assumed that in our last equation) the method of going around the perimeter of the garden and continuing to go around the same perimeter as it goes smaller until you get to the center of the lawn is actually equally as fast!
The equation for this second method is pretty easy to get once you take in a few smaller cases
n = 1; 1t, 0k (1
2 = 1 so it checks out)
n = 2; 2t [(1 + 1)t], 2k [(1 + 1)k] (2
2 = 4 so it checks out)
n = 3; 5t [(2 + 1 + 1 + 1)t], 4k [(1 + 1 + 1 + 1)k] (3
2 = 9 so it checks out)
n = 4; 10t [(3 + 2 + 2 + 1 + 1 + 1)t], 6k [(1 + 1 + 1 + 1 + 1 + 1)k] (4
2 = 16 so it checks out)
It may be hard to see, but you'll find that the amount of edges is given by the formula...
(n - 1) + 2(n - 2) + 2(n -3) + ... + 2 + 1.
The formula for the amount of corners is given by
(2n - 2)
So the formula for the perimeter side is actually
Equation 2) 2k(n - 1) + t((n - 1) + 2(n - 2) + 2(n - 3) + ... + 2 + 1)
Now we can easily see through inspection that both equation 1) and equation 2) are the same, because the amount of squares in the garden is equal, and because the equations for the corners are the same (they have the same amount of corners) then they must have the same amount of straight edges and thus must take the same amount of time to do.
However, if that proof does not satisfy you, we can prove it by induction
The assumption is that:
2k(n - 1) + t(n(n - 1) + 2) = 2k(n - 1) + t((n - 1) + 2(n - 2) + 2(n - 3) + ... + 2 + 1)
We can simplify the expression by getting rid of the corners (as they are the same 2k(n - 1) equation) and dividing both sides by "t", this gives us the final result that
n(n - 2) + 2 = (n - 1) + 2(n - 2) + 2(n - 3) + ... + 2 + 1
This is the result that we have to prove.
First of all, lets show that it's right for one case, let n = 1;
1(1 - 2) + 2 = (1 - 1) + 1;
1 = 1;
Thus it is true for n = 1;
Now, the method behind induction is that if we can prove it's true for the first case, can we prove it's true for the second case assuming the first case is true? Then prove it for the third case assuming the second case is true? Then the fourth case, then fifth, etc, etc, etc. So for this we assume that the xth case is true. So let n = x (normally you let n = k, but because we already used k we will use x, remember that x is a constant positive integer)
x(x - 2) + 2 = (x - 1) + 2(x - 2) + 2(x - 3) + ... + 2 + 1.
Now, we can assume the xth case is true, all we have to do is show that if the xth case is true, then the (x + 1)th case is true as well.
Consider n = x + 1;
(x + 1)(x - 1) + 2 = x + 2(x - 1) + 2(x - 3) + ... + 2 + 1;
All we have to do is show that the RHS is equal to the LHS,
RHS = x + 2(x - 1) + 2(x - 2) + ... + 2 + 1
Which could be written as...
RHS = x + (x - 1) + (x - 1) + 2(x - 2) + ... + 2 + 1;
Now remember we're assuming that the xth case is true, and the xth case states that...
x(x - 2) + 2 = (x - 1) + 2(x - 2) + 2(x - 3) + ... + 2 + 1.
So if we transfer that identity into our RHS equation we get
RHS = x + (x - 1) + x(x - 2) + 2
Now expand the RHS
RHS = x + x - 1 + x
2 - 2x + 2
RHS = x
2 + 1
That "+1" is giving us problems, but we can get rid of it by knowing that 1 = 2 - 1;
RHS = x
2 - 1 + 2
Now we can factorize using standard factorization formula
RHS = (x + 1)(x - 1) + 2
This is the left hand side
So therefore we have shown that the two statements are equivalent.
This is the full answer to the mathematics equation, that there are two fastest ways, the line method (the one you stated) and the perimeter method. Though you did give more thought than most people so thank you for actually considering the maths equations