If $1 + x + x^{2} + . . . . x^{p} = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8})$
then find the value of $p$

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Solution

$L . H . S . = \frac{1 - x^{p + 1}}{1 - x} = R . H . S .$
$∴ 1 - x^{p + 1} = (1 - x) (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8})$
$= (1 - x^{2}) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) = 1 - x^{16}$
$∴ p + 1 = 16$ $∴ p = 15$ .

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1 + x + x^{2} + . . . x^{p} = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8})

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