$1 + 2^{1} + 2^{2} + 2^{3} . . . . . . . . . {.2}^{1999}$ =

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Solution

The correct option is B

The given series is a GP

$1 + 2^{1} + 2^{2} + 2^{3} . . . . . . . . . {.2}^{1999} = \frac{2^{2000} - 1}{2 - 1}$

= $2^{2000} - 1$

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Similar questions

$1 + 2^{1} + 2^{2} + 2^{3} . . . . . . . . . {.2}^{1999}$ =

The remainder when $1 + 2^{1} + 2^{2} + 2^{3} . . . . . . . . . . . {.2}^{1999}$ is divided by 5 is:

1 + 2 + 2^{2} + 2^{3} + . . . . + 2^{1999}

5

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