what is the coefficient of xy^2
in the expansion of (x+y)^3?

Step-by-step explanation:

Using Binomial Expansion,

(x + y)³

= 3C0 * x³ + 3C1 * x²y + 3C2 * xy² + 3C3 * y³.

Therefore the coefficient of xy² is 3C2 = 3.

Answer:

The coefficient of the xy² term is 3.

For your question about xy⁵, the coefficient is 6

And for the x²y² question, it is also 6

Step-by-step explanation:

We can simply expand the expression to find the answer:

(x + y)³

= (x + y)(x + y)(x + y)

= (x + y)(x² + 2xy + y²)

= x³ + 2x²y + xy² + x²y + 2xy² + y³

= x³ + 3x²y + 3xy² + y³

So the term term containing xy² has a coefficient of 3

Now, there's actually a simple way to do this, and that's using Pascal's triangle.  If you're not familiar with it, here's a quick representation of the first few lines:

0                                    1

1                                  1     1

2                              1     2     1

3                           1     3     3     1

4                       1      4     6     4     1

5                    1      5    10    10    5    1

6                1      6     15   20   15    6    1

7             1      7     21   35   35   21   7    1

8         1      8     28   56  70   56   28  8   1

You'll notice that each number is simply the sum of the two numbers above it; below 4 and 6 wes see 10, below 20 and 15 is 35, etc.

This is very useful in this case, as the you may notice that the solution to each of the questions asked can be answered by this triangle.  For instance, when we solved (x + y)³, the coefficients of the terms were  1, 3, 3, and 1.  For a perfect square, it's 1, 2 and 1, and so on.  So to get the answers to your questions, you need only look at this triangle.

For example, want the the coefficent of the fourth term in the expansion of (x + y)⁷, it's easy, just look at the seventh row, and look for the fourth number, which would be 35.

But as we're given the x and y powers, but not the position of the term, we need to work that out as well.  Thankfully there's an easy trick to that.  The exponents will always be like this

[tex]x^{n}y^0 + x^{n - 1}y^1 + x^{n - 2}y^2 + ..... + x^2y^{n - 2} + x^1y^{n - 1} + x^0y^{n}[/tex]

where n is the number we're raising the binomial to.

We could put together an equation to solve this, but instead of going through a whole lesson, I'll just update the answer at the top.

The ratio of 6:5 can be expressed as

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