Regarding which variables get $k$ and which get $9-k$ in the expansion of $(x+3)^9$

Regarding



QUESTION

Hi sir in this question why can you swap the position of the $3$ and the $x$? because when Bec and I were trying to do it now, we put the power of $(9-k)$ on the $x$ term....​

ANSWER

Good question. The coefficients are totally symmetric just like Pascals Triangle, and can be seen in that rule we have proved before, namely


$\Large{n\choose k}={n\choose n-k}$


Now I am going to write the expansion of $\displaystyle (x+3)^9$ in two ways. They are equivalent and it doesn't matter which one you use, the answer is the same. I would suggest you put the $k$ on the more complicated expression as this makes algebra easier.

$$\normalsize\displaystyle (x+3)^9=\sum_{k=0}^n {n\choose k}x^{9-k}\, 3^k = \sum_{k=0}^n {n\choose k}x^{k}\,3^{9-k}$$

If you expand either version out fully the two expansions will be identical. The $k$ is just a dummy variable.