The sum of the series $1 + 2.2 + 3. 2^{2} + 4. 2^{4} + 5. 2^{4} + . . . + 100. 2^{99}$ is

99. 2^{99} + 1

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100. 2^{100}

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99. 2^{100}

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99. 2^{100} + 1

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Solution

The correct option is C $99. 2^{100} + 1$
Let $S = 1 + 2.2 + 3. 2^{2} + 4. 2^{4} + 5. 2^{4} + . . . + 100. 2^{99}$

$∴ 2 S = 1.2 + 2. 2^{2} + 3. 2^{3} + . . . + 99. 2^{99} + 100. 2^{100}$

Subtracting, we get

$- S = 1 + 1.2 + 1. 2^{2} + . . . + 1. 2^{99} - 100. 2^{100}$
$= (1 + 2 + 2^{2} + . . . + 2^{99}) - 100. 2^{100}$

$= \frac{1 (2^{100} - 1)}{2 - 1} - 100. 2^{100} = 2^{100} - 1 - 100. 2^{100}$
$∴ S = 100. 2^{100} - 2^{100} + 1 = 99. 2^{100} + 1$

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Similar questions

1 + 2.2 + {3.2}^{2} + {4.2}^{3} + {5.2}^{4} + \dots + {100.2}^{99}

1 + 2.2 + {3.2}^{2} + {4.2}^{3} + . . + {100.2}^{99}

Find the sum of the series $1.2 + {2.2}^{2} + {3.2}^{2} + . . . . . .100 {.2}^{100}$

1 + 2.2 + {3.2}^{2} + . . . . . + 100. 2^{99}

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