## Thomas' Calculus 13th Edition

$1+3x+6x^2+10x^3$
Apply Binomial series formula to find the first four terms. $(1+x)^m=1+\Sigma_{k=1}^\infty \dbinom{m}{k}x^k$ and $\dbinom{m}{k}=\dfrac{m(m-1)(m-2).....(m-k+1)}{k!}$ Thus,$(1-x)^{-3}=1+(-3)(-x)+\dfrac{(-3)(-4)(-x^2)}{2!}+\dfrac{(-3)(-4)(-5)(-x^3)}{3!}+...=1+3x+6x^2+10x^3+...$ Hence, we have the first four terms are: $1+3x+6x^2+10x^3$