Answer
$x^{32}+16x^{30}+120x^{28}$
Work Step by Step
Apply Binomial Theorem or Binomial expansion to find the first three terms of $(x^2+1)^{16}$.
$(x+y)^n=\displaystyle \binom{n}{0}x^ny^0+\displaystyle \binom{n}{1}x^{n-1}y^1+........+\displaystyle \binom{n}{n}x^0y^n$
Now, $(x^2+1)^{16}=\displaystyle \binom{16}{0}(x^2)^{16}(1)^0+\displaystyle \binom{16}{1}(x^2)^{15}(1)^1+\displaystyle \binom{16}{2}(x^2)^{14}(1)^2$
$=(1)x^{32}(1)+16(x^{30})(1)+120(x^{28})(1)$
Thus, $(x^2+1)^{16}=x^{32}+16x^{30}+120x^{28}$
