How to find $a$ and $n$ in the following equation $(1+ax)^n$ by using Pascal triangle?

Question

I have the following equation: $(1+ax)^n$ for which the first three terms are $1+24x+240 x^2$. I found the coefficients to be $a=4$ and $n=6$ by using the binomial expansion but can I do it by using the Pascal triangle?

Answer 1

First of all, the solution $n=6,a=4$ is correct.

I will assume $n$ is an integer. The terms $24x$ and $240x^2$ tell you the lines to look at in Pascal triangle are the $m$-th lines where $\binom{m}{1}a=24$ and $\binom{m}{2}a^2=240$. Then you get the system $$ \begin{cases} ma=24\\ m(m-1)a^2=480, \end{cases} $$ which gives the solution you found. Note that these are exactly the same steps you go through when you use the binomial expansion (indeed Binomial theorem and Pascal's triangle are intimately related).

Answer 2

Form the Pascal's triangle: $$\begin{array}{c|cc} n&1&24&240&\cdots&&&\\ \hline 1&1&a&&&&\\ 2&1&2a&a^2&&&\\ 3&1&3a&3a^2&a^3\\ 4&1&4a&6a^2&4a^3&a^4\\ 5&1&5a&10a^2&10a^3&5a^4&a^5\\ 6&1&6a&15a^2&20a^3&15a^4&6a^5&a^6\\ \vdots \end{array}$$ Now you must quickly go through the cases: $$\begin{align}2a&=24 \Rightarrow a=12 \Rightarrow a^2\ne 240\\ 3a&=24 \Rightarrow a=8 \Rightarrow 3a^2\ne 240\\ 4a&=24 \Rightarrow a=6 \Rightarrow 6a^2\ne 240\\ 5a&=24 \Rightarrow a=4.8 \Rightarrow 10a^2\ne 240\\ 6a&=24 \Rightarrow \color{red}{a=4} \Rightarrow 15a^2= 240 \Rightarrow \color{red}{n=6}\\ \end{align}$$