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

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

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Answer 1 · 2017-02-16T05:00:47

$a = 180$ .

Explanation:

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

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

$t_(k + 1) = color(white)(two)_nC_ka^(n - k)b^k$

We have

$k + 1 = 3$

$k = 2$

Use the formula now.

$t_3 = color(white)(two)_10C_2x^(10 -2)(-2y)^2$

The value of $color(white)(two)_10C_2$ can be computed using the formula $color(white)(two)_nC_r = (n!)/((n - r)!r!)$ . Therefore, $color(white)(two)_10C_2 = (10!)/(8!2!) = 45$

$t_3 = 45x^8 4y^2$

$t_3 = 180x^8y^2$

Therefore, $a= 180$ .

Hopefully this helps!

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

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