Solution:

794

The nth term in the series is given by the sum of x^n for x in the range 1 to 3, i.e. 1^n + 2^n + 3^n.

Thus the first term is 1^1 + 2^1 + 3^1 = 1 + 2 + 3 = 6,
the second term is 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14,
and so on.

The sixth term is then 1^6 + 2^6 + 3^6 = 1 + 64 + 729 = 794.

Solution:

BACK

All the others can also be read backwards as proper words.

Solution:

Each word has 3 consecutive letters of the alphabet-
couGHIng (GHI)
thiRSTy (RST)
DEFiant (DEF)

Solution:

SQRT(.2 ^ -2)

Take .2 and raise it to the power of -2 and then take the square root.

Solution:

HighJacking (High+Jack+King)

Solution:

Back To Square one

Solution:

Stand up and be counted

Solution:

No time at all it is already built

Solution:

6526884.

To get from 4 to 12, you use the following formula. 12 = (4 squared)/2 + 4.
Continue on with (X^2)/2 + X.

Solution:

As per the puzzle given above The fewest number of aircraft is 3! Imagine 3 aircraft (A, B and C). A is going to fly round the world. All three aircraft start at the same time in the same direction. After 1/6 of the circumference, B passes 1/3 of its fuel to C and returns home, where it is refueled and starts immediately again to follow A and C.

C continues to fly alongside A until they are 1/4 of the distance around the world. At this point C completely fills the tank of A which is now able to fly to a point 3/4 of the way around the world. C has now only 1/3 of its full fuel capacity left, not enough to get back to the home base. But the first 'auxiliary' aircraft reaches it in time in order to refuel it, and both 'auxiliary' aircraft are the able to return safely to the home base.

Now in the same manner as before both B and C fully refueled fly towards A. Again B refuels C and returns home to be refueled. C reaches A at the point where it has flown 3/4 around the world. All 3 aircraft can safely return to the home base, if the refueled process is applied analogously as for the first phase of the flight.