(1) 初項 a=1a = 1a=1 公比 r=0.2r = 0.2r=0.2 から収束 Sn=a1−r=11−0.2=54S_n = \frac{a}{1-r} = \frac{1}{1-0.2} = \frac{5}{4}Sn=1−ra=1−0.21=45 (2) a=1a = 1a=1 r=−3r = -\sqrt{3}r=−3 ∣r∣>1|r| > 1∣r∣>1 より発散 (3) a=6a = 6a=6 r=−13r = -\dfrac{1}{\sqrt{3}}r=−31 から収束 Sn=61−(−13)=633+1=18−633−1=9−33S_n = \frac{6}{1-\left(-\dfrac{1}{\sqrt{3}}\right)} = \frac{6\sqrt{3}}{\sqrt{3}+1} = \frac{18-6\sqrt{3}}{3-1} = 9-3\sqrt{3}Sn=1−(−31)6=3+163=3−118−63=9−33 (4) a=2+1a = \sqrt{2}+1a=2+1 r=2−12+1=3−22r = \dfrac{\sqrt{2}-1}{\sqrt{2}+1} = 3-2\sqrt{2}r=2+12−1=3−22 Sn=2+11−(3−22)=2+12(2−1)=(2+1)22(2−1)(2+1)=3+222=2+32S_n = \frac{\sqrt{2}+1}{1-(3-2\sqrt{2})} = \frac{\sqrt{2}+1}{2(\sqrt{2}-1)} = \frac{(\sqrt{2}+1)^2}{2(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{3+2\sqrt{2}}{2} = \sqrt{2}+\frac{3}{2}Sn=1−(3−22)2+1=2(2−1)2+1=2(2−1)(2+1)(2+1)2=23+22=2+23