Question

Find the sum of n terms and also Sum of S till infinity

S=(3/5)+(5/15)+(7/45)+(9/135)​

Answer 1

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Class 11

Sequence and Series

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$s = \frac{3}{5} + \frac{5}{15} + \frac{7}{45} + \frac{9}{135} ........... \infty \\ \\ \frac{s}{3} = \frac{3}{15} + \frac{5}{45} + \frac{7}{135} ................. \infty \\ \\ \frac{2s}{3} = \frac{3}{5} + \frac{2}{15} + \frac{2}{45} .................. \infty \\ \\ \frac{2s}{3} = \frac{3}{5} + \frac{2}{15} (1 + \frac{1}{3} + \frac{1}{ {3}^{2} }) \\ now \: we \: know \: that \: sum \: of \: infinity \: is \: \frac{a}{1 - r} \\ \\ therefore \\ \frac{2s}{3} = \frac{3}{5} + \frac{2}{15} \times ( \frac{1}{1 - \frac{1}{3} } ) \\ \frac{2s}{3} = \frac{3}{5} + \frac{2}{15} ( \frac{3}{2} ) \\ \frac{2s}{3} = \frac{3}{5} + \frac{1}{5} \\ \frac{s}{3} = \frac{2}{5} \\ s \infty = \frac{6}{5}$

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For sum of n terms

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$\frac{2S}{3}=\frac{3}{5}+\frac{2}{15}(1+\frac{1}{3}+\frac{1}{\{3}^{2}}........)-\frac{2n-1}{5\times \{3}^{n}} \\ \\ \frac{2s}{3}=\frac{3}{5}+\frac {2}{15} +(\frac{\{(\frac{1}{3})}^{n-1}{\frac{1}{3} -1} - """"""""" \\ \\ \frac{2s}{3}=\frac{3}{5}+\frac{1-\frac{1}{{3}^{n-1}}}{5}- """""""""""" \\ \\ S=\frac{9}{10}+\frac{3}{10}+(1-\frac{1}{{3}^{n-1}})-"""""""""""$

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Answer 2

Answer:

its 6/5

Step-by-step explanation:

hope u got ur ans