Solution: Suppose the farmer takes X pears, then:

The first customer bought half of all pears and added 1/2 to get the result.

Buy now: x ×1/2+1/2 = (x+1)/2.

Remaining pears: x-(x+1)/2 = (x-1)/2.

The second guest bought the remaining half of the pears and added 1/2.

Buy now: (x-1)/2×1/2+1/2 = (x+1)/4.

The remaining pears: (x-1)/2-(x+1)/4 = (x-3)/4.

The third guest bought half of the remaining pears and got 1/2.

Buy now: (x-3)/4×1/2+1/2 = (x+1)/8.

The remaining pears: (x-3)/4 -(x+ 1)/8 =(x-7)/8.

The fourth guest bought half of the remaining pears and got 1/2.

Buy now: (x-7)/8×1/2+1/2 = (x+1)16.

The remaining pears: (x-7)/8-(x+1)16 = (x-15)/16.

The fifth guest bought half of the remaining pears and got 1/2.

Buy now: (x-15)/16×1/2+1/2 = (x+1)/32.

Remaining pears: (x-15)/16-(x+1)/32 = (x-31)/32.

The sixth guest also bought the remaining half of the pears and got 1/2.

Buy now: (x-31)/32×1/2+1/2 = (x+1)/64.

The remaining pears: (x-31)/32-(x+1)/64 = (x-63)/64.

At this time, the farmer's pears had just been sold out.

That is, (x-63)/64=0 gives x = 63.

So, the farmer brought 63 pears.

Attachment (4) Standard problem-solving ideas: (ingenious methods)

Solution: Suppose the farmer takes X pears, then:

The first customer bought half of all pears and added 1/2 to get the result.

Buy now: x ×1/2+1/2 = (x+1)/2.

Remaining pears: x-(x+1)/2 = (x-1)/2.

It can be written as [x-(2- 1 1 power) ]/2 to 1 power.

The second guest bought the remaining half of the pears and added 1/2.

Buy now: (x-1)/2×1/2+1/2 = (x+1)/4.

The remaining pears: (x-1)/2-(x+1)/4 = (x-3)/4.

It can be written as [x-(2 squared-1)]/2 squared.

The third guest bought half of the remaining pears and got 1/2.

Buy now: (x-3)/4×1/2+1/2 = (x+1)/8.

The remaining pears: (x-3)/4 -(x+ 1)/8 =(x-7)/8.

It can be written as [x-(cubic of 2-1)]/cubic of 2.

By analogy, we can draw the following conclusions:

The nth guest bought half of the remaining pears and 1/2.

The remaining pears: [x-(n power of 2-1)]/n power of 2.

The nth guest just bought it,

That is, [x-(2 to the nth power-1)]/2 to the nth power = 0 gives x = 2 to the nth power-1.

Because the sixth guest just bought it, the farmer's pear is 2 to the sixth power-1, which is 63 (pieces)

So, the farmer brought 63 pears.

Summary: Attached (4) standard problem-solving ideas, it is more correct to summarize the formula according to observation and thinking first. If there are several pears in the title, the last question is not the sixth one, but the guests of 16 (or above 26th) also bought half of the remaining pears and 1/2. At this time, the farmer's pears had just been sold out. How many pears did the farmer bring? It is obviously too slow and complicated to solve general problems with Annex (4).