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

Oct 12, 2017

$128 {a}^{7} - 448 {a}^{6} b + 672 {a}^{5} {b}^{2} - 560 {a}^{4} {b}^{3} + 280 {a}^{3} {b}^{4} - 64 {a}^{2} {b}^{5} + 14 a {b}^{6} - {b}^{7}$

#### Explanation:

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:

$1 , 7 , 21 , 35 , 35 , 21 , 7 , \mathmr{and} 1$

The two terms in the binomial are: $2 a \mathmr{and} - b$

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

So:

$1 {\left(2 a\right)}^{7} {\left(- b\right)}^{0} + 7 {\left(2 a\right)}^{6} {\left(- b\right)}^{1} + 21 {\left(2 a\right)}^{5} {\left(- b\right)}^{2} + 35 {\left(2 a\right)}^{4} {\left(- b\right)}^{3} + 35 {\left(2 a\right)}^{3} {\left(- b\right)}^{4} + 21 {\left(2 a\right)}^{2} {\left(- b\right)}^{5} + 7 {\left(2 a\right)}^{1} {\left(- b\right)}^{6} + 1 {\left(2 a\right)}^{0} {\left(- b\right)}^{7}$

Now we have to simplify:

$128 {a}^{7} - 7 \left(64\right) {a}^{6} b + 21 \left(32\right) {a}^{5} {b}^{2} - 35 \left(16\right) {a}^{4} {b}^{3} + 35 \left(8\right) {a}^{3} {b}^{4} - 64 {a}^{2} {b}^{5} + 14 a {b}^{6} - {b}^{7}$

$128 {a}^{7} - 448 {a}^{6} b + 672 {a}^{5} {b}^{2} - 560 {a}^{4} {b}^{3} + 280 {a}^{3} {b}^{4} - 64 {a}^{2} {b}^{5} + 14 a {b}^{6} - {b}^{7}$