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Question Number 125638 by aurpeyz last updated on 12/Dec/20

Answered by mathmax by abdo last updated on 12/Dec/20

$$\mathrm{S}=\frac{3}{3}-\frac{5}{5^2}+\frac{7}{5^3}-\frac{9}{5^4}+...\Rightarrow \mathrm{S}=1+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{n}}}{5^{\mathrm{n}+1}}$$$${\mathrm{a}}_{\mathrm{n}}=\mathrm{x}\mathrm{n}+\mathrm{y}{\mathrm{a}}_1=5=\mathrm{x}+\mathrm{y}\mathrm{and}{\mathrm{a}}_2=7=2\mathrm{x}+\mathrm{y}\Rightarrow \{\begin{array}{l}\mathrm{x}+\mathrm{y}=5\\ 2\mathrm{x}+\mathrm{y}=7\end{array}$$$$\Rightarrow \mathrm{x}=2\Rightarrow \mathrm{y}=3\Rightarrow {\mathrm{a}}_{\mathrm{n}}=2\mathrm{n}+3\mathrm{we}\mathrm{get}{\mathrm{a}}_3=9...\Rightarrow$$$$\mathrm{S}=1+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}\frac{2\mathrm{n}+3}{5^{\mathrm{n}+1}}\mathrm{so}\mathrm{the}{\mathrm{n}}^{\mathrm{eme}}\mathrm{term}\mathrm{is}{(-1)}^{\mathrm{n}}\times \frac{2\mathrm{n}+3}{5^{\mathrm{n}+1}}$$

Commented by aurpeyz last updated on 13/Dec/20

$$thanks$$

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