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

1 Answer
Oct 12, 2017

Answer:

#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):
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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#