What is the fourth term in the expansion of (2x-y)^5?

Question

The answer is –40x^2y^3 but I have absolutely no clue how they got it.. and I have to show work. Please help!

Answer 1

See binomial theorem , in wikipedia.

The expansion of (a+b) ^ 5

=

1 (a^5) + 5 (a^4)( b^1) + 10 (a^3)(b^2) + 10 (a^2)(b^3) + 5(a^1)(b^4) + 1 (b^5)


where the coefficients 1,5,10,10,5,1 is the sixth line of "pascals triangle"

Here a = 2x
here b = -1y
So the fourth term is 10 [(2x)^2][(-y)^3] =

-40 x^2 y^3

done

Answer 2

Using Pascal's Triangle, we can find that the expansion in general form would be

(x + y)5 = x^5 +5x^4y + 10x^3y^2 +10x^2y^3 +5xy^4 + y^5

(look up Pascal's Triangle if you haven't seen it - I'm guessing you either haven't or aren't used to using it, have it memorized if you want to make life simpler) and, by substituting into the fourth term of the general form

10x^2y^3

our solution is

10(2x)^2(-1y)^3 = 10*4*-1*x^2y^3 = -40x^2y^3.

Answer 3

By the binomial theorem, it is
C(5, 3) (2x)^(5-3) (-y)^3
where C(5, 3) = 5 choose 3 = 10
so it is
10 (2^2) x^2 (-1)^3 y^3 = -40x^2y^3.

In general the (k+1)'th term in the expansion of (a+b)^n is
C(n, k) a^(n-k) b^k. Here n = 5, k = 3, a = (2x), b = (-y).

Answer 4

let a = 2x and b = -y
(a+b)^5

Use pascal's triangle or binomial theorem, everybody else did binomial theorem, so pascal's triangle

n=0 1
n=1 1 1
n=2 1 2 1
n=3 1 3 3 1
n=4 1 4 6 4 1
n=5 1 51010 5 1 (a+b)^5

coefficient of fourth term is 10

10a^2b^3
but a is 2x and b is -y
10[(2x)^2]*(-y)^3
-40x^2 y^3

Answer 5

Use the binomial theorem. (5 taken 3) * (2x)^2 * (-y)^3.
5 taken 3 is 5! / (3! * 2!) which is 10...
So the answer is -40x^2y^3

Answer 6

the nth term is: 5Cn . (2x)^n. (-y)^(5-n), for n=0, 1, ..., 5