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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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Find the sum of the GP $ 1, 2, 4, 8, 16, 64 .... $ upto n terms

$(a)\;n^{2} - 1\qquad(b)\;256\qquad(c)\;1 - 2^{n} \qquad(d)\;2^{n} - 1$

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  • Sum of n terms of a GP is $S_{n} = \large \frac{a ( r^{n} - 1)} {r - 1} $
Sum of n terms of a GP is $S_{n} = \large \frac{a ( r^{n} - 1)} {r - 1} $
Here $ a = 1, r = 2 $
Sum $S_{n} = \large \frac {1 (2^{n} - 1) } {2 - 1} $
$S_{n} = 2^{n} - 1 $
answered Nov 19, 2013 by harini.tutor
 

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