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

$1259712 {x}^{2} {y}^{7}$
The binomial theorem states that (a+b)^n=sum_(r=0)^n""(""_r^n)a^(n-r)b^r. In this question, $a = 4 x$, $b = 3 y$, and $n = 9$.
The eight term occurs when $r = 8 - 1 = 7$ (since we sum starting with $r = 0$).
We just need to substitute these values in (""_r^n)a^(n-r)b^(r) to get (""_7^9)(4x)^(9-7)(3y)^7=36*16x^2*2187*y^7=1259712x^2y^7.