If $1 + b + b^{2} + . . . + b^{2} + . . . . + b^{n} = (1 + b) (1 + b^{2}) (1 + b^{4}) (1 + b^{8})$ then n is given by

7

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9

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15

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17

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Solution

The correct option is C $15$
Given
$\frac{b^{n + 1} - 1}{b - 1} = (1 + b) (1 + b^{2}) (1 + b^{4}) (1 + b^{8})$

$\Rightarrow b^{n + 1} - 1 = (b^{2} - 1) (1 + b^{2}) (1 + b^{4}) (1 + b^{8}) = b^{16} - 1$

$∴ n = 15$

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If 1+b+b2+...+b2+....+bn=(1+b)(1+b2)(1+b4)(1+b8) then n is given by

The correct option is C 15Givenbn+1−1b−1=(1+b)(1+b2)(1+b4)(1+b8) ⇒bn+1−1=(b2−1)(1+b2)(1+b4)(1+b8)=b16−1∴n=15

If $1 + b + b^{2} + . . . + b^{2} + . . . . + b^{n} = (1 + b) (1 + b^{2}) (1 + b^{4}) (1 + b^{8})$ then n is given by

The correct option is C $15$
Given
$\frac{b^{n + 1} - 1}{b - 1} = (1 + b) (1 + b^{2}) (1 + b^{4}) (1 + b^{8})$

$\Rightarrow b^{n + 1} - 1 = (b^{2} - 1) (1 + b^{2}) (1 + b^{4}) (1 + b^{8}) = b^{16} - 1$

$∴ n = 15$