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Question Number 157454 by tounghoungko last updated on 23/Oct/21

$$Given1+\frac{3}{y}+\frac{5}{y^2}+\frac{7}{y^3}+\frac{9}{y^4}+...=71$$$$then1+\frac{4}{y}+\frac{9}{y^2}+\frac{16}{y^3}+\frac{25}{y^4}+...=?$$

Answered by FongXD last updated on 23/Oct/21

$$\mathrm{let}1+\frac{3}{\mathrm{y}}+\frac{5}{{\mathrm{y}}^2}+\frac{7}{{\mathrm{y}}^3}+\frac{9}{{\mathrm{y}}^4}+...=71\mathrm{be}\mathrm{equation}\left(1\right)$$$$\mathrm{and}1+\frac{4}{\mathrm{y}}+\frac{9}{{\mathrm{y}}^2}+\frac{16}{{\mathrm{y}}^3}+\frac{25}{{\mathrm{y}}^4}+...=\mathrm{M}\mathrm{be}\mathrm{equation}\left(2\right)$$$$\mathrm{take}\left(2\right)-\left(1\right):(1+\frac{4}{\mathrm{y}}+\frac{9}{{\mathrm{y}}^2}+\frac{16}{{\mathrm{y}}^3}+\frac{25}{{\mathrm{y}}^4}+...)-(1+\frac{3}{\mathrm{y}}+\frac{5}{{\mathrm{y}}^2}+\frac{7}{{\mathrm{y}}^3}+\frac{9}{{\mathrm{y}}^4}+...)=\mathrm{M}-71$$$$\iff \mathrm{M}-71=\frac{1}{\mathrm{y}}+\frac{4}{{\mathrm{y}}^2}+\frac{9}{{\mathrm{y}}^3}+\frac{16}{{\mathrm{y}}^4}+\frac{25}{{\mathrm{y}}^5}+...$$$$\iff \mathrm{M}-71=\frac{\mathrm{M}}{\mathrm{y}},\Rightarrow \mathrm{M}=\frac{71\mathrm{y}}{\mathrm{y}-1}$$$$\mathrm{we}\mathrm{have}\left(1\right):1+\frac{3}{\mathrm{y}}+\frac{5}{{\mathrm{y}}^2}+\frac{7}{{\mathrm{y}}^3}+\frac{9}{{\mathrm{y}}^4}+...=71$$$$\iff (1+\frac{1}{\mathrm{y}}+\frac{1}{{\mathrm{y}}^2}+...)+2(\frac{1}{\mathrm{y}}+\frac{1}{{\mathrm{y}}^2}+\frac{1}{{\mathrm{y}}^3}+...)+2(\frac{1}{{\mathrm{y}}^2}+\frac{1}{{\mathrm{y}}^3}+\frac{1}{{\mathrm{y}}^4}+...)+...=71$$$$\iff \frac{1}{1-\frac{1}{\mathrm{y}}}+\frac{2\cdot \frac{1}{\mathrm{y}}}{1-\frac{1}{\mathrm{y}}}+\frac{2\cdot \frac{1}{{\mathrm{y}}^2}}{1-\frac{1}{\mathrm{y}}}+\frac{2\cdot \frac{1}{{\mathrm{y}}^3}}{1-\frac{1}{\mathrm{y}}}+...=71$$$$\iff -1+2(1+\frac{1}{\mathrm{y}}+\frac{1}{{\mathrm{y}}^2}+\frac{1}{{\mathrm{y}}^3}+...)=71(1-\frac{1}{\mathrm{y}})$$$$\iff -1+\frac{2}{1-\frac{1}{\mathrm{y}}}=71(1-\frac{1}{\mathrm{y}})$$$$\iff 71{(1-\frac{1}{\mathrm{y}})}^2+(1+\frac{1}{\mathrm{y}})-2=0$$$$\iff 1-\frac{1}{\mathrm{y}}=\frac{-1+\sqrt{569}}{142},[\because 0<\frac{1}{\mathrm{y}}<1]$$$$\Rightarrow \mathrm{y}=\frac{142}{143-\sqrt{569}}$$$$\mathrm{we}\mathrm{get}:\mathrm{M}=\frac{71\left(\frac{142}{143-\sqrt{569}}\right)}{\frac{142}{143-\sqrt{569}}-1}=\frac{71\times 142}{142-(143-\sqrt{569})}=\frac{10082}{\sqrt{569}-1}$$$$\Rightarrow 1+\frac{4}{\mathrm{y}}+\frac{9}{{\mathrm{y}}^2}+\frac{16}{{\mathrm{y}}^3}+\frac{25}{{\mathrm{y}}^4}+...=\frac{5041(\sqrt{569}+1)}{284}$$

Commented by Tawa11 last updated on 23/Oct/21

$$\mathrm{Great}\mathrm{sir}$$

Commented by FongXD last updated on 23/Oct/21

$$\mathrm{Thanks},\mathrm{sir}$$

Answered by qaz last updated on 23/Oct/21

$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{2\mathrm{n}+1}{{\mathrm{y}}^{\mathrm{n}}},(\mid \frac{1}{\mathrm{y}}\mid <1)$$$$=(2\mathrm{xD}+1){\mid }_{\mathrm{x}=1/\mathrm{y}}\frac{1}{1-\mathrm{x}}$$$$={\left[\frac{1+\mathrm{x}}{{(1-\mathrm{x})}^2}\right]}_{\mathrm{x}=1/\mathrm{y}}$$$$=\frac{\mathrm{y}(1+\mathrm{y})}{{(1-\mathrm{y})}^2}=71$$$$\Rightarrow \frac{1}{\mathrm{y}}=\frac{140}{143+\sqrt{569}}$$$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(\mathrm{n}+1)}^2}{{\mathrm{y}}^{\mathrm{n}}}$$$$=({\mathrm{x}}^2{\mathrm{D}}^2+3\mathrm{xD}+1){\mid }_{\mathrm{x}=1/\mathrm{y}}\frac{1}{1-\mathrm{x}}$$$$={[\frac{{2\mathrm{x}}^2}{{(1-\mathrm{x})}^3}+\frac{3\mathrm{x}}{{(1-\mathrm{x})}^2}+\frac{1}{1-\mathrm{x}}]}_{\mathrm{x}=1/\mathrm{y}}$$$$={[\frac{{2\mathrm{x}}^2}{{(1-\mathrm{x})}^3}+\frac{\mathrm{x}}{{(1-\mathrm{x})}^2}+\frac{1+\mathrm{x}}{{(1-\mathrm{x})}^2}]}_{\mathrm{x}=1/\mathrm{y}}$$$$={\left[\frac{{\mathrm{x}}^2+\mathrm{x}}{{(1-\mathrm{x})}^3}\right]}_{\mathrm{x}=1/\mathrm{y}}+71$$$$=71{\left[\frac{\mathrm{x}}{1-\mathrm{x}}\right]}_{\mathrm{x}=1/\mathrm{y}}+71$$$$=\frac{71}{4}(\sqrt{569}+1)$$

Commented by tounghoungko last updated on 24/Oct/21

$$thx$$

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