22) Determine o coeficiente \(a_{98}\) na expansão de:
$$sin(2x+\frac{\pi}{4}) = \Sigma_{n=0}^{\infty}a_nx^n$$
Usando:\(\sin(x+y) = \sin(y)\cos(x)+\cos(y)\sin(x)\), obtém-se:
$$sin(2x+\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\left(\sin(2x) + cos(2x) \right)$$
Visto que a expansão de sin(x) e cos(x) são respectivamente:
$$\sum^{\infty}_{n=0} (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
e
$$\sum^{\infty}_{n=0} (-1)^n\frac{x^{2n}}{(2n)!}$$
Aplicando na equação:
$$\sin(2x+\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\left( \underbrace{\sin(2x)}_{\sum^{\infty}_{n=0}(-1)^n\frac{(2x)^{2n+1}}{(2n+1)!}} + \underbrace{cos(2x)}_{\sum^{\infty}_{n=0} (-1)^n\frac{(2x)^{2n}}{(2n)!}}\right) =
\frac{\sqrt{2}}{2}\left( {\sum^{\infty}_{n=0}(-1)^n\frac{(2x)^{2n+1}}{(2n+1)} + {\sum^{\infty}_{n=0} (-1)^n\frac{(2x)^{2n}}{(2n)!}}}\right)$$
$$ = \frac{\sqrt{2}}{2} \sum^{\infty}_{n=0}(-1)^n\left[\frac{(2x)^{(2n+1)}}{(2n+1)!} + \frac{(2x)^{2n}}{(2n)!}\right]$$
$$ = \frac{\sqrt{2}}{2} \sum^{\infty}_{n=0}{-1}^{\lfloor\frac{n}{2}\rfloor}\frac{2^n}{n!}x^n$$
Obtendo \(a_{98}\):
$$a_{98} = \frac{\sqrt{2}}{2}\cdot(-1)\frac{2^{98}}{98!} = - \sqrt{2}\frac{2^{97}}{98!}$$