How do you find the 8th term in the expansion of the binomial ${(4 x + 3 y)}^{9}$ $(4x+3y)^9$ ?

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How do you find the 8th term in the expansion of the binomial ${(4 x + 3 y)}^{9}$ $(4x+3y)^9$ ?

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Answer 1 · 2017-04-02T13:19:20

$1259712 x^{2} y^{7}$ $1259712x^2y^7$

Explanation:

The binomial theorem states that ${(a + b)}^{n} = n \sum r = 0 (_{r}^{n}) a^{n - r} b^{r}$ $(a+b)^n=sum_(r=0)^n""(""_r^n)a^(n-r)b^r$ . In this question, $a = 4 x$ $a=4x$ , $b = 3 y$ $b=3y$ , and $n = 9$ $n=9$ .

The eight term occurs when $r = 8 - 1 = 7$ $r=8-1=7$ (since we sum starting with $r = 0$ $r=0$ ).

We just need to substitute these values in $(_{r}^{n}) a^{n - r} b^{r}$ $(""_r^n)a^(n-r)b^(r)$ to get $(_{7}^{9}) {(4 x)}^{9 - 7} {(3 y)}^{7} = 36 \cdot 16 x^{2} \cdot 2187 \cdot y^{7} = 1259712 x^{2} y^{7}$ $(""_7^9)(4x)^(9-7)(3y)^7=36*16x^2*2187*y^7=1259712x^2y^7$ .

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How do you find the 8th term in the expansion of the binomial ${(4 x + 3 y)}^{9}$ $(4x+3y)^9$ ?

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How do you find the 8th term in the expansion of the binomial (4x+3y)9(4x+3y)^9?

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How do you find the 8th term in the expansion of the binomial ${(4 x + 3 y)}^{9}$ $(4x+3y)^9$ ?