4)$\displaystyle{\because F_{n+1}-F_{n-1}=F_n \therefore \frac{1}{F_{n-1}F_{n+1}}=\frac{1}{F_n} \left[ \frac{1}{F_{n-1}}-\frac{1}{F_{n+1}} \right]}$
$ \displaystyle{\sum_{n=2}^{\infty}\frac{1}{F_{n-1}F_{n+1}}= \sum_{n=2}^{\infty}\frac{1}{F_n}\left[ \frac{1}{F_{n-1}}-\frac{1}{F_{n+1}} \right]}$
$\displaystyle{=\left[\frac{1}{F_{1}F_{2}}-\frac{1}{F_{2}F_{3}}\right]+\left[\frac{1}{F_{2}F_{3}}-\frac{1}{F_{3}F_{4}}\right]+\cdots=\frac{1}{F_{1}F_{2}}=1}$
ต่อด้วยข้อ5เลยละกันนะครับ
5)$Evaluate\;the\;series\;\displaystyle{\frac{1}{6}+\frac{1}{36}+\frac{3}{216}+\frac{17}{1296}+\frac{83}{7776}+\cdots}$
ขออภัยครับงั้นผมขอเปลี่ยนใหม่นะครับ
5)จงหา3พจน์ถัดไปของลำดับ$\longrightarrow1, 11, 21, 1211, 111221,\cdots$
__________________
$$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x-b\sin x}{a\sin x+b\cos x}dx=\ln\left(\frac{a}{b}\right)$$
BUT
$$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x+b\sin x}{a\sin x+b\cos x}dx=\frac{\pi ab}{a^{2}+b^{2}}+\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\ln\left(\frac{a}{b}\right)$$