![\bf ~~~~~~~~\textit{binomial theorem expansion} \\\\ (x+3)^4\implies \begin{array}{cccl} term&coefficient&value\\ \cline{1-3}&\\ 1&+1&(x)^4(+3)^0\\ 2&+4&(x)^3(+3)^1\\ 3&+6&(x)^2(+3)^2\\ 4&+4&(x)^1(+3)^3\\ 5&+1&(x)^0(+3)^4 \end{array} \\\\\\ 1(x)^4(3)^0+4(x)^3(3)^1+(x)^2(3)^2+4(x)^1(3)^3+(x)^0(3)^4 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill x^4+12x^3+9x^2+108x+81~\hfill](https://img.qammunity.org/2020/formulas/mathematics/middle-school/j2v0f9tb9qfetn72v3508uawg3uk8al2jh.png)
how did I get the coefficient of 6 for the 3rd expanded term?
by multiplying 4*3, and then dividing that by 2.
the 4(current coefficient), the 3(x's current exponent), and the 2(3's exponent on the following term).
so in essence, the product of the current coefficient by the exponent of the first term divided by the exponent of the second term in the following term.
now, you can always find the nth factor in any binomial by
![\bf \stackrel{\stackrel{coefficient}{\downarrow }}{\left(\cfrac{n!}{k!(n-k)!}\right)} \qquad \stackrel{\stackrel{\textit{first term factor}}{\downarrow }}{\left( a^(n-k) \right)} \qquad \stackrel{\stackrel{\textit{second term factor}}{\downarrow }}{\left( b^k \right)}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/uysdtgtsckkngiyavwxywp8p4qk73hipis.png)
where "k" is the term, say you want the 51st term? then k = 50.
and "n" is the exponent of the binomial.