Question

Find the sum of the following series to infinity:

(i) $1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}+...\infty$

(ii) 8 + $4\sqrt{2}$ + 4 + ... ∞

(iii) 2/5 + 3/5² +2/5³ + 3/5⁴ + ... ∞.

(iv) 10 − 9 + 8.1 − 7.29 + ... ∞

(v) $\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{56}+...\infty$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{In}\mathrm{the}\mathrm{given}\mathrm{G}.\mathrm{P}.,\mathrm{first}\mathrm{term},a=1 \\ \mathrm{and}\mathrm{common}\mathrm{ratio},r=-\frac{1}{3} \\ \mathrm{Hence},\mathrm{the}\mathrm{sum}S\mathrm{to}\mathrm{infinity}\mathrm{is}\mathrm{given}\mathrm{by}S=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{1}{1-\left(-{\displaystyle \frac{1}{3}}\right)}=\frac{3}{4}. \\ \left(\mathrm{ii}\right)\mathrm{In}\mathrm{the}\mathrm{given}\mathrm{G}.\mathrm{P}.,\mathrm{first}\mathrm{term},a=8 \\ \mathrm{and}\mathrm{common}\mathrm{ratio},\mathit{}r=\frac{1}{\sqrt{2}} \\ \mathrm{Hence},\mathrm{the}\mathrm{sum}S\mathit{}\mathrm{to}\mathrm{infinity}\mathrm{is}\mathrm{given}\mathrm{by}S=\frac{a}{1-r}=\frac{8}{1-{\displaystyle \frac{1}{\sqrt{2}}}}=\left(2+\sqrt{2}\right). \\ \left(\mathrm{iii}\right)\mathrm{We}\mathrm{have}: \\ \frac{2}{5}+\frac{3}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}. \\ =\left(\frac{2}{5}+\frac{2}{5^3}+..\infty \right)+\left(\frac{3}{5^2}+\frac{3}{5^4}+...\right) \\ =\left(\mathrm{An}\mathrm{infinite}\mathrm{G}.\mathrm{P}.\mathrm{with}\mathit{}a=\frac{2}{5}\mathrm{and}\mathit{}r=\frac{1}{25}\right)+\left(\mathrm{An}\mathrm{infinite}\mathrm{G}.\mathrm{P}.\mathrm{with}\mathit{}a=\frac{3}{25}\mathrm{and}r=\frac{1}{25}\right) \\ =\left(\frac{{\displaystyle \frac{2}{5}}}{1-{\displaystyle \frac{1}{25}}}\right)+\left(\frac{{\displaystyle \frac{3}{5}}}{1-{\displaystyle \frac{1}{25}}}\right) \\ =\left(\frac{{\displaystyle \frac{2}{5}}}{{\displaystyle \frac{24}{25}}}\right)+\left(\frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{24}{25}}}\right) \\ =\left(\frac{10}{24}+\frac{3}{24}\right) \\ =\frac{13}{24} \end{gathered}$$