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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Sequence and Series

Sum of the series $\;\large\frac{2}{5}+\large\frac{3}{5^2}+\large\frac{4}{5^3}+\large\frac{2}{5^4}+\large\frac{3}{5^5}+\large\frac{4}{5^6}+...$ equals:

$(a)\;\large\frac{71}{124}\qquad(b)\;\large\frac{69}{124}\qquad(c)\;\large\frac{35}{124}\qquad(d)\;\large\frac{49}{124}$

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1 Answer

Answer : (b) $\;\large\frac{69}{124}$
Explanation : $\;S=\large\frac{2}{5}+\large\frac{3}{5^2}+\large\frac{4}{5^3}+\large\frac{2}{5^4}+\large\frac{3}{5^5}+\large\frac{4}{5^6}+...$
$\large\frac{1}{5^3}$$S=\large\frac{2}{5^4}+\large\frac{3}{5^5}+\large\frac{4}{5^6}+...$
Subtracting ,
$(1-\large\frac{1}{5^3})$$S=\large\frac{2}{5}+\large\frac{3}{5^2}+\large\frac{4}{5^3}$
$\large\frac{124}{125}\;$$S=\large\frac{50+15+4}{125}$
$S=\large\frac{69}{124}\;.$
answered Jan 23, 2014 by yamini.v
edited Mar 21, 2014 by balaji
 

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