How do you find the 3rd term of the expansion of (2x-y)^5

[suzi]

How do you find the 3rd term of the expansion of (2x-y)^5

 

[AKG]

Have you learned anything on Binomial Theorem or Pascal's Triangle?
Basically:

$$(x+a)^n = \sum_{k=0}^n {n\choose k} x^k a^{n-k}$$

Replace:
"x" with "2x"
"a" with "-y"
"n" with "5"

Since the summation goes from k=0, the third term would be when k=2. So the third term is:

$${5\choose 2} (2x)^2 (-y)^{5-2}$$
$$= -40x^2y^3$$

In case you don't understand what ${n\choose k}$ is, it is simply:

$$\frac{n!}{k!(n-k)!}$$

If you don't want to calculate what ${n\choose k}$ is, you can construct Pascal's Triangle:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
. . .
. . .

If you write it out yourself so it take the shape of an equilateral triangle, not a right one like I have drawn, something like this:

      1
    1  1
  1  2  1
1  3  3  1

Then you'll notice that each number is the sum of the two numbers above it. So to find ${n\choose k}$, you look at the $(k+1)^{th}$ element in the $(n+1}^{th}$ row. This is actually probably a lot more annoying than actually calculating it. Most calculators have a "choose" operation, look for something that might say "nCr". Also, you can do it easily in you head a lot of the time. Basically, if you're doing 5 choose 2, for example, instead of finding 5! by multiplying 5x4x3x2x1, just multiply the first 2, i.e. 5x4. Then, divide by 2!. If it were 10 choose 3, then only do the first 3, i.e. 10x9x8, then divide by 3!, which is just 6. Finding 2, 3, and 4 factorial are pretty easy, and it's pretty easy to multiply the first 3 or four numbers, even if you're getting numbers as big as 12, 12x11x10 is easy. Moreover, you can see if you're doing 12 choose 3, and easily see that 3! is 6, you can simply calculate (12/6)x11x10 = 2x11x10 = 220, which was really easy.