# 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.