Question

How do you use the binomial series to expand `(2a-b)^7`?

Answer 1

`128a^7-448a^6b+672a^5b^2-560a^4b^3+280a^3b^4-64a^2b^5+14ab^6-b^7`

Explanation:

The binomial theorem shows up in Pascal’s triangle (start with 1’s and then add adjacent numbers to find the number below):

Since we are taking this binomial to the 7th power, we have to look at the 7th row (where the single 1 on top is considered our 0th row) to find our coefficients.

Our coefficients are:

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

The two terms in the binomial are: `2a and -b`

I have to multiply these terms together with the coefficients so that `2a` has a power descending from 7 to 0 and `-b` has a power ascending from 0 to 7.

So:

`1(2a)^7(-b)^0+7(2a)^6(-b)^1+21(2a)^5(-b)^2+35(2a)^4(-b)^3+35(2a)^3(-b)^4+21(2a)^2(-b)^5+7(2a)^1(-b)^6+1(2a)^0(-b)^7`

Now we have to simplify:

`128a^7-7(64)a^6b+21(32)a^5b^2-35(16)a^4b^3+35(8)a^3b^4-64a^2b^5+14ab^6-b^7`

`128a^7-448a^6b+672a^5b^2-560a^4b^3+280a^3b^4-64a^2b^5+14ab^6-b^7`