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

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

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AKG

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Basically:

[tex](x+a)^n = \sum_{k=0}^n {n\choose k} x^k a^{n-k}[/tex]

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:

[tex]{5\choose 2} (2x)^2 (-y)^{5-2}[/tex]

[tex]= -40x^2y^3[/tex]

In case you don't understand what [itex]{n\choose k}[/itex] is, it is simply:

[tex]\frac{n!}{k!(n-k)!}[/tex]

If you don't want to calculate what [itex]{n\choose k}[/itex] 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:

Code:

```
1
1 1
1 2 1
1 3 3 1
```

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