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

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Solution

The correct option is A
0

The given sequence is Geometric progression with first term 1 and common ratio 2.

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

= $2^{2000} - 1$

$2^{2000} - 1 = 4^{1000} - 1 = (5 - 1)^{1000} - 1$

$(5 - 1)^{1000} - 1$ $= [^{1000} C_{0} (5)^{1000} -^{1000} C_{1} (5)^{999} +^{1000} C_{2} (5)^{998} . . . . . . . . -^{1000} C_{999} (5) +^{1000} C_{1000}] - 1$

$(5 - 1)^{1000} - 1$ $= [^{1000} C_{0} (5)^{1000} -^{1000} C_{1} (5)^{999} +^{1000} C_{2} (5)^{998} . . . . . . . . -^{1000} C_{999} (5) + 1] - 1$

$(5 - 1)^{1000} - 1$ $=^{1000} C_{0} (5)^{1000} -^{1000} C_{1} (5)^{999} +^{1000} C_{2} (5)^{998} . . . . . . . . -^{1000} C_{999} (5)$

$(5 - 1)^{1000} - 1$ = A multiple of 5

$\Rightarrow$ The remainder will be zero.

Suggest Corrections

Similar questions

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

1 + 2 + 2^{2} + 2^{3} + \dots + 2^{1999}

5

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

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