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Question Number 195129 by a.lgnaoui last updated on 25/Jul/23

$$\mathrm{Calculer}\mathrm{la}\mathrm{valeur}\mathrm{de}\mathrm{la}\mathrm{serie}\mathrm{suivante}:$$$$\mathbf{S}=\frac{3}{2}+\frac{5}{8}+\frac{7}{32}+\frac{9}{128}+.....$$

Answered by MM42 last updated on 25/Jul/23

$$\frac{1}{2^2}S=\frac{3}{2^3}+\frac{5}{2^5}+\frac{7}{2^7}+...$$$$\Rightarrow (1-\frac{1}{4})S=\frac{3}{2}+2(\frac{1}{2^3}+\frac{1}{2^5}+...)$$$$=\frac{3}{2}+2\left(\frac{\frac{1}{8}}{1-\frac{1}{4}}\right)=\frac{3}{2}+\frac{1}{3}=\frac{11}{6}✓$$

Commented by a.lgnaoui last updated on 25/Jul/23

$$\mathrm{thanks}$$

Commented by a.lgnaoui last updated on 25/Jul/23

$$\mathrm{but}\mathrm{the}\mathrm{answer}\mathrm{is}\mathrm{wrong}$$

Commented by MM42 last updated on 25/Jul/23

$$yes$$$$\frac{3}{4}S=\frac{11}{6}\Rightarrow S=\frac{11}{6}\times \frac{4}{3}=\frac{22}{9}$$

Answered by BaliramKumar last updated on 25/Jul/23

$$\mathbf{S}=\frac{3}{2}+\frac{5}{8}+\frac{7}{32}+\frac{9}{128}+.....\left(\mathrm{i}\right)$$$$\frac{\mathrm{S}}{4}=\frac{3}{8}+\frac{5}{32}+\frac{7}{128}+...........\left(\mathrm{ii}\right)$$$$\left(\mathrm{i}\right)-\left(\mathrm{ii}\right)$$$$\mathrm{S}-\frac{\mathrm{S}}{4}=\frac{3}{2}+(\frac{5}{8}-\frac{3}{8})+(\frac{7}{32}-\frac{5}{32})+(\frac{9}{128}-\frac{7}{128})+.......$$$$\frac{3\mathrm{S}}{4}=\frac{3}{2}+\left(\frac{1}{4}\right)+\left(\frac{1}{16}\right)+\left(\frac{1}{64}\right)+.......$$$$\frac{3\mathrm{S}}{4}=\frac{3}{2}+\frac{\frac{1}{4}}{1-\frac{1}{4}}$$$$\frac{3\mathrm{S}}{4}=\frac{3}{2}+\frac{1}{3}=\frac{11}{6}$$$$\mathrm{S}=\frac{11\times 4}{6\times 3}$$$$\mathrm{S}=\frac{22}{9}$$$$$$$$$$

Commented by a.lgnaoui last updated on 25/Jul/23

$$\mathrm{Exact},\mathrm{thanks}$$

Answered by a.lgnaoui last updated on 25/Jul/23

$$4\mathrm{S}=6+\frac{5}{2}+\frac{7}{8}+\frac{9}{32}+...$$$$4\mathrm{S}-\mathrm{S}=6+1+\frac{1}{4}+\frac{1}{16}+...$$$$3\mathrm{S}=6+(1+\frac{1}{4}+\frac{1}{4^2}+....)$$$$=6+\frac{1}{1-\frac{1}{4}}=6+\frac{4}{3}=\frac{22}{3}$$$$\Rightarrow \mathbf{S}=\frac{22}{9}$$

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