If the sum of $1 + 2 + 2^{2} + . . . . . . . . . . . . . . . . + 2^{n - 1}$ is 255. find the value of n.

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Solution

$G P : 1 + 2 + 2^{2} + . . . . . . . . . . . . . . . . + 2^{n - 1} First term, a = 1 Common ratio, r = \frac{2^{2}}{2} = 2 Given, Sum = 255 Sum of GP = \frac{a (r^{n} - 1)}{r - 1} 255 = \frac{1 (2^{n} - 1)}{2 - 1} 255 = 2^{n} - 1 2^{n} = 256 2^{n} = 2^{8} n = 8$

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