How to find the fourth term of a binomial expansion?

So let's look at how to expand a binomial..
Start with just the variable exponents which will be numbered in ascending and descending order.

example (without coefficients)
(x+y)^3 = x^3(y^0) + x^2(y^1) + x^1(y^2) + x^0(y^3)

If the binomial is (x-y)^3 the +/- signs will toggle like so.. (again leaving out coefficients)
(x-y)^3 = x^3(y^0) - x^2(y^1) + x^1(y^2) - x^0(y^3)

If there is a coefficient or power inside the binomial like (x^2+2y)^3 these are also raised to the appropriate power.

(x^2+2y)^3 = (x^2)^3(y^0)(2^0) + (x^2)^2(y^1)(2^1) +

now there are also coefficients for each term which follow a combination pattern of Pascal triangle. C(3,0) ; C(3,1) ; C(3,2) ; C(3,3)
these get multiplied by each term.

So the 4th term of (x^2-2y)^3
will be negative
have coefficient of: C(3,3) * (2^3) = 8
variables: x^0(y^3) = y^3
= -8y^3

a. geometricb. g[h] = (3/4)·g[h-1]; g[1] = 384c. g[h] = 384·(3/4)^(h-1)d. g[12] ≈ 16.2183 gallonse. g[12] ≈ 64.8732 quarts

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt