Find the coefficient of $\sqrt{3}$ in $(1+\sqrt{3})^7$? I just want to ask you if my solution is correct.
Here's the problem,
Using the Binomial Theorem, find the coefficient of $\sqrt{3}$ in $(1+\sqrt{3})^7$.
Solution: The binomial theorem is,
$$\sum_{r=0}^n{{n}\choose{r}}x^{n-r}y^r=(x+y)^n,$$Hence, by letting x = 1 and y = $\sqrt{3}$, then we have$$\sum_{r=0}^7{{7}\choose{r}}1^{7-r}{\sqrt{3}}^r=(1+\sqrt{3})^7,$$And to take the coefficient of $\sqrt{3}$ we need to set r = 1. That is,$${{7}\choose{1}}\sqrt{3}=(1+\sqrt{3})^7$$Hence, the coefficient of $\sqrt{3}$ is ${{7}\choose{1}}=7\qquad \blacksquare$
Is that correct?

Applying the binomial theorem,
$(1+√3)^7 =1+ 7×√3  + 21×3 +  35×3√3+  35×9 + 21×9√3  +7×27  + 1×27√3$
hence coefficient of $√3$ is
   $$7 + 35×3 + 21×9 + 27=328.$$
