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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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