Answer
$1+3x^2+3x^4+x^6$
Work Step by Step
Apply the binomial series formula to determine the first four terms:
$(1+x)^r=1+\Sigma_{k=1}^\infty \dbinom{r}{k}x^k$
Here, we have $\dbinom{r}{k}=\dfrac{r(r-1)(r-2).....(r-k+1)}{k!}$
Now, we have $(1+x^2)^{3}=1+3x^2+\dfrac{(3)(2)}{2!}(x^2)^2+\dfrac{(3)(2)(1)}{3!}(x^2)^3+0$
Hence, our first four terms are: $1+3x^2+3x^4+x^6$
