Answer
$x^{34}+17x^{32}+136x^{30}$
Work Step by Step
Apply Binomial Theorem or Binomial expansion to find the first three terms of $(x^2+1)^{17}$.
$(x+y)^n=\displaystyle \binom{n}{0}x^ny^0+\displaystyle \binom{n}{1}x^{n-1}y^1+........+\displaystyle \binom{n}{n}x^0y^n$
Now, $(x^2+1)^{17}=\displaystyle \binom{17}{0}(x^2)^{17}(1)^0+\displaystyle \binom{17}{1}(x^2)^{16}(1)^1+\displaystyle \binom{17}{2}(x^2)^{15}(1)^2$
$=(1)x^{34}(1)+17(x^{32})(1)+136x^{30}$
Thus, $(x^2+1)^{17}=x^{34}+17x^{32}+136x^{30}$
