Question

Find the sum of the following geometric series:
(i) 0.15 + 0.015 + 0.0015 + ... to 8 terms;

(ii) $\sqrt{2}+\frac{1}{\sqrt{2}}+\frac{1}{2\sqrt{2}}+...\mathrm{to}8\mathrm{terms};$

(iii) $\frac{2}{9}-\frac{1}{3}+\frac{1}{2}-\frac{3}{4}+...\mathrm{to}5\mathrm{terms};$

(iv) (x +y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + ... to n terms;

(v) $\frac{3}{5}+\frac{4}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+...\mathrm{to}2n\mathrm{terms};$

(vi) $\frac{a}{1+i}+\frac{a}{(1+i{)}^2}+\frac{a}{(1+i{)}^3}+...+\frac{a}{(1+i{)}^n}.$

(vii) 1, −a, a², −a³, ... to n terms (a ≠ 1)

(viii) x³, x⁵, x⁷, ... to n terms

(ix) $\sqrt{7},\sqrt{21},3\sqrt{7},...\mathrm{to}n\mathrm{terms}$

Answer 1

(i) Here, a = 0.15 and r $=\frac{a_2}{a_1}=\frac{0.015}{0.15}=\frac{1}{10}$.
$$\begin{gathered} {\mathrm{S}}_8=a\left(\frac{1-r^8}{1-r}\right) \\ =0.15\left(\frac{1-{\left({\displaystyle \frac{1}{10}}\right)}^8}{1-{\displaystyle \frac{1}{10}}}\right) \\ =0.15\left(\frac{1-{\displaystyle \frac{1}{{10}^8}}}{{\displaystyle \frac{1}{10}}}\right) \\ =\frac{1}{6}\left(1-\frac{1}{{10}^8}\right) \end{gathered}$$

(ii) Here, a = $\sqrt{2}$ and r = $\frac{1}{2}$.
$$\begin{gathered} {\mathrm{S}}_8=\mathrm{a}\left(\frac{1-{\mathrm{r}}^8}{1-\mathrm{r}}\right) \\ =\sqrt{2}\left(\frac{1-{\left({\displaystyle \frac{1}{2}}\right)}^8}{1-{\displaystyle \frac{1}{2}}}\right) \\ =\sqrt{2}\left(\frac{1-{\displaystyle \frac{1}{256}}}{{\displaystyle \frac{1}{2}}}\right) \\ =2\sqrt{2}\left(\frac{255}{256}\right) \\ =\frac{255\sqrt{2}}{128} \end{gathered}$$

(iii) Here, a =$\frac{2}{9}\mathrm{and}r=-\frac{3}{2}$.
$$\begin{gathered} S_5=a\left(\frac{r^5-1}{r-1}\right) \\ =\frac{2}{9}\left(\frac{{\left({\displaystyle \frac{-3}{2}}\right)}^5-1}{{\displaystyle \frac{-3}{2}}-1}\right) \\ =\frac{2}{9}\left(\frac{\left(-{\displaystyle \frac{243}{32}}\right)-1}{{\displaystyle \frac{-3}{2}}-1}\right) \\ =\frac{2}{9}\left(\frac{{\displaystyle \frac{-275}{32}}}{{\displaystyle \frac{-5}{2}}}\right) \\ =\frac{1100}{1440} \\ =\frac{55}{72} \end{gathered}$$