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

Let a_{n} = 10 \times {(\frac{1}{\sqrt{2}})}^{n}

a_{n} is a geometrical sequence

S_{n} = a_{0} + a_{1} + \dots + a_{n - 1}

S_{n} = 10 \times \frac{1 - {(\frac{1}{\sqrt{2}})}^{n}}{1 - (\frac{1}{\sqrt{2}})}

Proove that :

S_{n} = \frac{10 \sqrt{2}}{\sqrt{2} - 1} \times (1 - {(\frac{1}{\sqrt{2}})}^{n})

Answered by Kunal12588 last updated on 22/Apr/19

S_{n} = 10 \times \frac{1 - {(\frac{1}{\sqrt{2}})}^{n}}{1 - \frac{1}{\sqrt{2}}} [G i v e n]

= 10 \sqrt{2} \times \frac{1 - {(\frac{1}{\sqrt{2}})}^{n}}{\sqrt{2} - 1}

= \frac{10 \sqrt{2}}{\sqrt{2} - 1} (1 - {(\frac{1}{\sqrt{2}})}^{n})

w e l l W h a t w a s t h e q u e s t i o n ?

Commented by Hassen_Timol last updated on 22/Apr/19

Thank you sir

Answered by tanmay last updated on 22/Apr/19

a_{n} = 10 \times {(\frac{1}{\sqrt{2}})}^{n} \to a_{1} = 10 {(\frac{1}{\sqrt{2}})}^{1}

a_{2} = 10 \times {(\frac{1}{\sqrt{2}})}^{2}

f o r m u l a S_{n} = \frac{a (1 - r^{n})}{1 - r}

r = \frac{a_{2}}{a_{1}} = \frac{10 \times {(\frac{1}{\sqrt{2}})}^{2}}{10 \times {(\frac{1}{\sqrt{2}})}^{1}} = \frac{1}{\sqrt{2}}

S_{n} = \frac{a (1 - r^{n})}{1 - r}

= \frac{10 \times \frac{1}{\sqrt{2}} {(1 - {(\frac{1}{\sqrt{2}})}^{n}}}{1 - \frac{1}{\sqrt{2}}}

= \frac{10 \times {(1 - {(\frac{1}{\sqrt{2}})}^{n}}}{\sqrt{2} - 1}

Commented by Hassen_Timol last updated on 22/Apr/19

Thank you Sir