## Precalculus (6th Edition) Blitzer

$x^{32}+16x^{30}+120x^{28}$
Calculate the first three terms of $(x^2+1)^{16}$ by using Binomial Theorem or Binomial expansion. This implies $(x^2+1)^{16}=\displaystyle \binom{16}{0}(x^2)^{16}1^0+\displaystyle \binom{16}{1}(x^2)^{15}1^1+\displaystyle \binom{16}{2}(x^2)^{14}1^2$ or, $=x^{32}+16(x^{30})(1)+120x^{28}$ or, $=x^{32}+16x^{30}+120x^{28}$