Question

What is the number of terms of the expanded form of (x+3y)^7?

Answer 1

It has `8` terms, starting with `x^7` and ending with `(3y)^7 = 3^7y^7`

Explanation:

To expand a binomial raised to the `7`th power, use the row of Pascal's triangle that starts `1, 7, 21,...`. Some people call the first row of Pascal's triangle the `0`th row, which would make this the `7`th row. I prefer to call it the `8`th row.

Then

`(a+b)^7 = a^7 + 7a^6b + 21a^5b^2 + 35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7`

In our case `a = x` and `b = 3y`, so there are all those powers of `3` to deal with too.

Write out the row of Pascal's triangle:

`1, 7, 21, 35, 35, 21, 7, 1`

Write out ascending powers of `3` from `3^0 = 1` up to `3^7`:

`1, 3, 9, 27, 81, 243, 729, 2187`

Multiply the two sequences together:

`1, 21, 189, 945, 2835, 5103, 5103, 2187`

These are the coefficients we need:

`(x+3y)^7 = x^7 + 21x^6y + 189x^5y^2 + 945x^4y^3 + 2835x^3y^4 + 5103x^2y^5 + 5103xy^6 + 2187y^7`