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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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$\sqcap_{n=2}^{\infty}\;\large\frac{n^3-1}{n^3+1}$ equals

$(a)\;\large\frac{1}{3}\qquad(b)\;\large\frac{2}{3}\qquad(c)\;\large\frac{2}{7}\qquad(d)\;\large\frac{1}{10}$

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Answer : (d) $\;\large\frac{1}{10}$
Explanation : $\sqcap_{n=2}^{\infty}\;\large\frac{n^3-1}{n^3+1}$
$=\sqcap_{n=2}^{\infty}\;\large\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$
$=\large\frac{1}{3}(\large\frac{7}{3})\;.\large\frac{2}{4}(\large\frac{13}{7})\;.\large\frac{3}{5}(\large\frac{21}{13})$
$=\large\frac{2}{3}\;.$
answered Jan 23, 2014 by yamini.v
 

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