The question is about finding $n$, given that $$(5+nx)^2 (1+(3/5)x)^n = 25+100x+\cdots$$
I've tried 2 approaches.
Method 1:
$$5^2 (1+nx)^2$$
$$25(1+(2)(nx))$$
$$25+50nx$$
And on the other hand...
$$(1+ 3/5 x)^n$$
$$1+(3/5 x)(n)$$
Therefore
$$(25+50nx)(1+ 3/5 nx)$$
$$25+ 75/5 nx+50nx+\cdots $$
I haven't include past $x^1$ because in the question it doesn't mention anything further
$$25+15nx+50nx+\cdots$$
$$25+65nx+\cdots $$
Therefore
$$100x = 65nx,\quad \text{so} \quad 100/65 = n$$
However, the answer is $4$ in the answer booklet because of a $10nx+15nx$ to get $25nx$ and divide evenly into $100x$, however, I don't see where I went wrong and need help finding this
Upon reflection, I thought about using the regular binomial theorem formula, as seen in Method 2:
$$(5+nx)^2$$
$$(2nCr1)(5)^2 (nx)^1$$
etc, but I don't think it's the way to go because it doesn't make sense for $r$ to be anything else other than $1$ and that eliminates the possibility of having the constant of $25$
I'm confident in Method 1, but just need some help with finding where I went wrong.