Browse Questions

Find the sum of the GP $1, 2, 4, 8, 16, 64 ....$ upto n terms

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

Can you answer this question?

Toolbox:
• 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