<s>[INST] What is the $8^{th}$ term of $\left(3x-\frac{y}{2}\right)^{10}$? What is the $8^{th}$ term of $\left(3x-\frac{y}{2}\right)^{10}$?
My solution: I'am not sure if I'am correct :)
$^{10}C_r (3x)^{10-r} \left(-\frac{y}{2}\right)^r$
where $r= 7$ since we start at $r=0$
$^{10}C_7 (3x)^3 \left(-\frac{y}{2}\right)^7$
$\left(-120\right)(27x^3)\left(\frac{y^7}{128}\right)$
$=  -\left(\frac{405}{16}\right) x^3 y^7$
 [/INST] $$(3x-(1/2)y)^{10}$$ general term is $$T_k=\binom{n}{k}a^{n-k}b^k$$ in our case we have
$$a=3x,b=-\frac{1}{2}y,n=10,k=7$$
so
$$T_7=\binom{10}{7}(3x)^3(-\frac{1}{2}y)^7=-\frac{405}{16}x^3y^7$$
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