How do you find the coefficient of a of the term ax^6y^2 in the expansion of the binomial (2x-3y)^8?

May 7, 2017

Coefficient $a$ has a value of $16 , 128$.

Explanation:

This will be the $3$rd term of the expansion. This is because we always start $x$ at the highest power and $y$ at the lowest. Also, there will be $8 + 1 = 9$ terms in the expansion, so the term of the form $a {x}^{6} {y}^{2}$ will be the $9 - 6 = 3$rd term.

The formula ${t}_{k + 1} = {\textcolor{w h i t e}{t w o}}_{n} {C}_{k} {\left(x\right)}^{n - k} {y}^{k}$ will let you find the term that you're looking for of ${\left(x + y\right)}^{n}$.

$3 = k + 1 \to k = 2$

Now, substitute your known values into the formula.

${t}_{3} = {\textcolor{w h i t e}{t w o}}_{8} {C}_{2} {\left(2 x\right)}^{8 - 2} {\left(- 3 y\right)}^{2}$

We now notice that $a = 64 \cdot 9 \cdot {\textcolor{w h i t e}{t w o}}_{8} {C}_{2}$. We can evaluate this using the combination formula, which states that color(white)(two)_nC_r = (n!)/(r!(n - r)!), where n! = n(n - 1)(n - 2)(...)(1).

a = (8!)/(2!(8 - 2)!) * 64 * 9

a = (8!)/(6!2!) * 64 * 9

a = (8 * 7 * 6!)/(6! * 2!) * 64 * 9

$a = \frac{56}{2} \cdot 64 \cdot 9$

$a = 28 \cdot 64 \cdot 9$

$a = 16 , 128$

Hopefully this helps!