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-

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

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

Answered by Kunal12588 last updated on 22/Apr/19

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

Commented by Hassen_Timol last updated on 22/Apr/19

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

Answered by tanmay last updated on 22/Apr/19

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

Commented by Hassen_Timol last updated on 22/Apr/19

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

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