Infinite Sum Pattern

$\begin{aligned} S & = 1 - \frac 12 x^2 + \frac 1{24}x^4 - \frac 1{720}x^6 + ... \\ & = \frac 1{0!}x^0 - \frac 1{2!} x^2 + \frac 1{4!}x^4 - \frac 1{6!}x^6 + ... \end{aligned}$

This implies that the next term is $\dfrac 1{8!}x^8 = \boxed{\dfrac 1{40320}x^8}$ and $\displaystyle S = \sum_{n=0}^\infty \frac {(-1)^n}{(2n)!}x^{2n}$ .

Notation: $!$ denotes the factorial notation . For example: $8!=1\times 2 \times 3 \times \cdots \times 8$ .

Looking at the denominators of consecutive terms, they follow a pattern of ${d_n} = {{(2n)}!}$ , ${n} = {0,1,2,...}$ . The ${x}$ terms also follow a similar pattern of ${x_{n}} = {x^{2n}}$ , ${n} = {0,1,2,...}$ . The signs also alternate between ${+}$ and ${-}$ . The term we're looking for is in the place ${n} = {4}$ , and the previous term is negative, so this one is positive. Plugging in ${n} = {4}$ , we get ${a_4} = \frac{1}{{(2*4)}!}*{x^{(2*4)}} = \frac{1}{8!}*{x^{8}}$ . Expanding the $8!$ , we receive the number $40320$ in the denominator, giving you $\frac{1}{40320}*{x^{8}}$