The positive integral values of n such that ${1.2}^{1} + {2.2}^{2} + {3.2}^{3} + {4.2}^{4} + {5.2}^{5} + . . . . . + n {.2}^{n} = 2^{(n + 10)} + 2 i s$

313

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513

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413

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613

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Solution

The correct option is B 513
$2^{1} + 2^{2} + 2^{3} + . . . + 2^{n} = 2^{n + 1} - 2$
$2^{2} + 2^{3} + . . {.2}^{n} = 2^{n + 1} - 2^{2}$
$2^{3} + . . {.2}^{n} = 2^{n + 1} - 2^{3}$
... ... ... ...
$+ 2^{n} = 2^{n + 1} - 2$
__________________
$= n (2^{n + 1}) - (2^{n + 1} - 2)$
$= 2^{n + 1} (n - 1) + 2$
Given that $2^{n + 1} (n - 1) + 2 = 2^{2 + 10} + 2$
$\Rightarrow (n - 1) 2^{n + 1} - 2^{n + 10}$
$\Rightarrow n - 1 = 2^{9}$
$\Rightarrow n = 2^{9} + 1 = 513$

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Similar questions

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2. 2^{2} + 3. 2^{3} + 4. 2^{4} + \dots + n . 2^{n} = 2^{n + 10}

$1.2 + {2.2}^{2} + {3.2}^{3} + \dots + n . 2^{n} = (n - 1) 2^{n + 1} + 2.$

$1.2 + {2.2}^{2} + {3.2}^{3} + . . . . + n {.2}^{n} = (n - 1) 2^{n + 1} + 2$

1.2 + {2.2}^{2} + {3.2}^{3} + . . . + n {.2}^{n} = (n - 1) 2^{n + 1} + 2.

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