How do you find the coefficient of $a$ $a$ of the term $a x^{8} y^{2}$ $ax^8y^2$ in the expansion of the binomial ${(x - 2 y)}^{10}$ $(x-2y)^10$ ?

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How do you find the coefficient of $a$ $a$ of the term $a x^{8} y^{2}$ $ax^8y^2$ in the expansion of the binomial ${(x - 2 y)}^{10}$ $(x-2y)^10$ ?

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Answer 1 · 2017-02-16T05:00:47

$a = 180$ $a = 180$ .

Explanation:

If we include the 0th term, then there will be 11 terms in this expansion. This means that the term $a x^{8} y^{2}$ $ax^8y^2$ will be the third term (since $11 - 8 = 3)$ $11- 8 = 3)$ .

The formula for the nth term in a binomial expansion ${(a + b)}^{n}$ $(a+ b)^n$ is given by

$t_{k + 1} = {t w o}_{n} C_{k} a^{n - k} b^{k}$ $t_(k + 1) = color(white)(two)_nC_ka^(n - k)b^k$

We have

$k + 1 = 3$ $k + 1 = 3$

$k = 2$ $k = 2$

Use the formula now.

$t_{3} = {t w o}_{10} C_{2} x^{10 - 2} {(- 2 y)}^{2}$ $t_3 = color(white)(two)_10C_2x^(10 -2)(-2y)^2$

The value of ${t w o}_{10} C_{2}$ $color(white)(two)_10C_2$ can be computed using the formula ${t w o}_{n} C_{r} = \frac{n!}{(n - r)! r!}$ $color(white)(two)_nC_r = (n!)/((n - r)!r!)$ . Therefore, ${t w o}_{10} C_{2} = \frac{10!}{8! 2!} = 45$ $color(white)(two)_10C_2 = (10!)/(8!2!) = 45$

$t_{3} = 45 x^{8} 4 y^{2}$ $t_3 = 45x^8 4y^2$

$t_{3} = 180 x^{8} y^{2}$ $t_3 = 180x^8y^2$

Therefore, $a = 180$ $a= 180$ .

Hopefully this helps!

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How do you find the coefficient of $a$ $a$ of the term $a x^{8} y^{2}$ $ax^8y^2$ in the expansion of the binomial ${(x - 2 y)}^{10}$ $(x-2y)^10$ ?

1 Answer

Explanation:

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How do you find the coefficient of $a$ $a$ of the term $a x^{8} y^{2}$ $ax^8y^2$ in the expansion of the binomial ${(x - 2 y)}^{10}$ $(x-2y)^10$ ?