By definition, the Binomial Theorem states:
$$(x+y)^n = {n\choose 0}x^n + {n\choose 1}x^{n-1}y + {n\choose 2}x^{n-2}y^2 + \cdots + {n \choose {n-1}}xy^{n-1} + {n \choose n}y^n$$
For any $x,y\in\mathbb R$ and $n \in \mathbb N$.
Now I know I could expand this simply by using the binomial theorem as stated above and just look at the coefficient of $x$ where $n = 7$, however that gets very lengthy, especially when – like in this case – our $n$ is a large number.
I was hoping for a better understanding through an explanation of how I may go about tackling this problem without expanding to the $n$th term.
Best Answer
In the expansion of $(a+x)^n$, the general term, that is the $(r+1)^{th}$ term is $$T_{r+1}=\binom nr a^{n-r}x^r.$$
Thus, in the expansion of $(1+x)^{23}$, the general term is $$\binom{23}r1^{23-r}x^r.$$
You want the coefficient of $x^7$, that is $r=7$, so the the coefficient is $$\binom{23}7.$$
This is the way, using general terms, you can derive any coefficient of any term without any expansion.