Let y ( x ) = ( 1 + x ) ( 1 + x 2 ) ( 1 + x 4 ) ( 1 + x 8 ) ( 1 + x 16 ) . Then y ' - y ' ' at x = - 1 is equal to [2023]

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Solution

Q.
Let $y (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})$ . Then $y' - y''$ at $x = - 1$ is equal to [2023]

1 976

2 944

3 496

4 464

Ans.
(3)

Given, $y (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})$

$= \frac{1 - x^{32}}{1 - x} = 1 + x + x^{2} + x^{3} + \dots + x^{31}$

$y' (x) = 1 + 2 x + 3 x^{2} + 4 x^{3} + \dots + 31 x^{30}$

$y' (- 1) = 1 - 2 + 3 - 4 + \dots - 30 + 31$

$= - 1 - 1 - 1 \dots (upto 15 times) + 31 = 16$

$y'' (x) = 2 + 6 x + 12 x^{2} + 20 x^{3} + \dots + 31 \times 30 x^{29}$

$y'' (- 1) = 2 - 6 + 12 - 20 + \dots - 31 \times 30$

$= - 4 - 8 - 12 - \dots (15 terms) = - 4 \times \frac{15 \times 16}{2} = - 480$

$y' - y'' = 16 + 480 = 496$

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