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Question Number 58364 by Hassen_Timol last updated on 22/Apr/19

Let a_n = 10 × ((1/(√2)))^n a_n is a geometrical sequence S_n = a_0 + a_1 + ... + a_(n−1) S_n = 10 × ((1 − ((1/(√2)))^n )/(1 − ((1/(√2))))) Proove that : S_n = ((10(√2))/((√2) − 1)) × (1−((1/(√2)))^n )

$$\mathrm{Let}\:{a}_{{n}} =\:\mathrm{10}\:×\:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{{n}} \\ $$$$\:\:\:\:{a}_{{n}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{geometrical}\:\mathrm{sequence} \\ $$$$\:\:{S}_{{n}} \:=\:{a}_{\mathrm{0}} \:+\:{a}_{\mathrm{1}} \:+\:...\:+\:{a}_{{n}−\mathrm{1}} \\ $$$$\:\:\:\:\:{S}_{{n}} =\:\mathrm{10}\:×\:\frac{\mathrm{1}\:−\:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{{n}} }{\mathrm{1}\:−\:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)} \\ $$$$\boldsymbol{\mathrm{Proove}}\:\boldsymbol{\mathrm{that}}\:: \\ $$$$ \\ $$$${S}_{{n}} \:=\:\frac{\mathrm{10}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}\:−\:\mathrm{1}}\:×\:\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{{n}} \right) \\ $$$$ \\ $$

Answered by Kunal12588 last updated on 22/Apr/19

S_n =10×((1−((1/(√2)))^n )/(1−(1/(√2)))) [Given] =10(√2)×((1−((1/(√2)))^n )/((√2)−1)) =((10(√2))/((√2)−1))(1−((1/(√2)))^n ) well What was the question?

$${S}_{{n}} =\mathrm{10}×\frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{{n}} }{\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}}\:\left[{Given}\right] \\ $$$$=\mathrm{10}\sqrt{\mathrm{2}}×\frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{{n}} }{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$$=\frac{\mathrm{10}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}−\mathrm{1}}\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{{n}} \right) \\ $$$${well}\:{What}\:{was}\:{the}\:{question}? \\ $$

Commented by Hassen_Timol last updated on 22/Apr/19

Thank you sir

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by tanmay last updated on 22/Apr/19

a_n =10×((1/(√2)))^n →a_1 =10((1/(√2)))^1 a_2 =10×((1/(√2)))^2 formula S_n =((a(1−r^n ))/(1−r)) r=(a_2 /a_1 )=((10×((1/(√2)))^2 )/(10×((1/(√2)))^1 ))=(1/(√2)) S_n =((a(1−r^n ))/(1−r)) =((10×(1/(√2)){(1−((1/(√2)))^n })/(1−(1/(√2)))) =((10×{(1−((1/(√2)))^n })/((√2) −1))

$${a}_{{n}} =\mathrm{10}×\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{{n}} \rightarrow{a}_{\mathrm{1}} =\mathrm{10}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{1}} \\ $$$${a}_{\mathrm{2}} =\mathrm{10}×\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$${formula}\:{S}_{{n}} =\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}} \\ $$$${r}=\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }=\frac{\mathrm{10}×\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }{\mathrm{10}×\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{1}} }=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$${S}_{{n}} =\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}} \\ $$$$=\frac{\mathrm{10}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left\{\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{{n}} \right\}\right.}{\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \\ $$$$=\frac{\mathrm{10}×\left\{\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{{n}} \right\}\right.}{\sqrt{\mathrm{2}}\:−\mathrm{1}} \\ $$

Commented by Hassen_Timol last updated on 22/Apr/19

Thank you Sir

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

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