# Binomial Expansion - Finding the term independent of n.

The coefficient of $x^2$ in the expansion of $\left(1 + \frac x5\right)^n$, where $n$ is a positive integer, is $\frac 35$ .

$(i)$ Find the value of $n$.

$(ii)$ Using this value of $n$, find the term independent of $x$ in the expansion of $\left(1 + \frac x5\right)^n \times \left(2- \frac 3x\right)^2$

For part $(i)$ I used the Binomial theorem and got the result where $n= 6$. Had no bigger issues with solving for $n$.

I do, however, struggle with part $(ii)$. I am not quite sure what I am supposed to use/do here. Do I also use the binomial theorem? I tried that and got nowhere since I do not know what "$r$" or "$n$" value to use since there are two different powers of binomial.

Any help would be appreciated.

For a general $n$, the constant term in $$f(x) = (1 + x/5)^n (2-3/x)^2$$ can be found by observing first that $$(2-3/x)^2 = 2 - 12x^{-1} + 9x^{-2}.$$ Then we see that the constant term of $f$ is given by $$1 \cdot 2 + \binom{n}{1} (x/5)(-12x^{-1}) + \binom{n}{2}(x/5)^2(9x^{-2}),$$ since these are the only terms for which the power of $x$ will be zero.