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Question Number 122608 by bramlexs22 last updated on 18/Nov/20

 Sum the series to 10^(th)  terms    (1/(1+(√x))) + (1/(1−x)) +(1/(1−(√x))) +...  equal to ___

$$\:{Sum}\:{the}\:{series}\:{to}\:\mathrm{10}^{{th}} \:{terms}\: \\ $$$$\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{x}}}\:+\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:+\frac{\mathrm{1}}{\mathrm{1}−\sqrt{{x}}}\:+... \\ $$$${equal}\:{to}\:\_\_\_\: \\ $$

Answered by liberty last updated on 18/Nov/20

 T_1  = ((1−(√x))/(1−x)) ; T_2 = (1/(1−x)) ; T_3 = ((1+(√x))/(1−x))   check  { ((T_2 −T_1 = ((√x)/(1−x)))),((T_3 −T_2 = ((√x)/(1−x)))) :} . it follows  that it is a AP , so S_(10) = 5(((2−2(√x))/(1−x)) + ((9(√x))/(1−x)))   S_(10)  = 5(((2+7(√x))/(1−x))) = ((10+35(√x))/(1−x))

$$\:\mathrm{T}_{\mathrm{1}} \:=\:\frac{\mathrm{1}−\sqrt{\mathrm{x}}}{\mathrm{1}−\mathrm{x}}\:;\:\mathrm{T}_{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\:;\:\mathrm{T}_{\mathrm{3}} =\:\frac{\mathrm{1}+\sqrt{\mathrm{x}}}{\mathrm{1}−\mathrm{x}} \\ $$$$\:\mathrm{check}\:\begin{cases}{\mathrm{T}_{\mathrm{2}} −\mathrm{T}_{\mathrm{1}} =\:\frac{\sqrt{\mathrm{x}}}{\mathrm{1}−\mathrm{x}}}\\{\mathrm{T}_{\mathrm{3}} −\mathrm{T}_{\mathrm{2}} =\:\frac{\sqrt{\mathrm{x}}}{\mathrm{1}−\mathrm{x}}}\end{cases}\:.\:\mathrm{it}\:\mathrm{follows} \\ $$$$\mathrm{that}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{AP}\:,\:\mathrm{so}\:\mathrm{S}_{\mathrm{10}} =\:\mathrm{5}\left(\frac{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{x}}}{\mathrm{1}−\mathrm{x}}\:+\:\frac{\mathrm{9}\sqrt{\mathrm{x}}}{\mathrm{1}−\mathrm{x}}\right) \\ $$$$\:\mathrm{S}_{\mathrm{10}} \:=\:\mathrm{5}\left(\frac{\mathrm{2}+\mathrm{7}\sqrt{\mathrm{x}}}{\mathrm{1}−\mathrm{x}}\right)\:=\:\frac{\mathrm{10}+\mathrm{35}\sqrt{\mathrm{x}}}{\mathrm{1}−\mathrm{x}} \\ $$

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