Answer:

Step-by-step explanation:
We will prove by mathematical induction that, for every natural n,

We will prove our base case (when n=1) to be true:
Base case:
As stated in the qustion, 
Inductive hypothesis:
Given a natural n,

Now, we will assume the inductive hypothesis and then use this assumption, involving n, to prove the statement for n + 1.
Inductive step:
Let´s analyze the problem with n+1 stones. In order to move the n+1 stones from A to C we have to:
- Move the first n stones from A to C (
moves). - Move the biggest stone from A to B (1 move).
- Move the first n stones from C to A (
moves). - Move the biggest stone from B to C (1 move).
- Move the first n stones from A to C (
moves).
Then,
.
Therefore, using the inductive hypothesis,

With this we have proved our statement to be true for n+1.
In conlusion, for every natural n,

Consider the picture.
Let MN be the midsegment of the trapezoid.
That is M is the midpoint of AD, N is the midpoint of BC.
Being the midsegment of the trapezoid, MN is parallel to the bases.
Let O and K be the intersections of the diagonals with the midsegment.
MN//AB, so MO//AB, and since M is the midpoint of DA, O must be the midpoint of DB,
Similarly we prove that K is the midpoint of CA.
Thus O is F and K is E.
O and K lie on the midsegment MN, so F and E lie on the midsegment.
MO is a midsegment of triangle ABD so |MO|=1/2 |AB|=1/2 * 10=5
MK is a midsegment of triangle ADC, so |MK|=1/2 * |DC|=1/2 * 22=11
|OK|=|MK|-|MO|=11-5=6 (units)
Answer:
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Step-by-step explanation:
Actually you can't solve this. There is no solutions.