If $\frac{d}{d x}\left[(x+1)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\right]$ $=\left(15 x^p-16 x^q+1\right)(x-1)^{-2}$, then $(p, q)$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{d}{d x}\left[(x+1)\left(x^2+1ight)\left(x^4+1ight)\left(x^8+1ight)ight] \ &=\frac{\left(15 x^ho-16 x^9+1ight)}{(x-1)^2} \ & 	ext { LHS }= \frac{d}{d x}\left[\frac{\left(x^2-1ight)\left(x^2+1ight)\left(x^4+1ight)\left(x^8+1ight)}{(x-1)}ight] \ &=\frac{d}{d x}\left[\frac{\left(x^4-1ight)\left(x^4+1ight)\left(x^8+1ight)}{(x-1)}ight] \ &=\frac{d}{d x}\left[\frac{\left(x^8-1ight)\left(x^8+1ight)}{(x-1)}ight] \ &=\frac{d}{d x}\left[\frac{\left(x^{16}-1ight)}{(x-1)}ight] \ &= \frac{(x-1)\left(16 x^{15}ight)-\left(x^{16}-1ight)}{(x-1)^2} \ &= \frac{16 x^{16}-16 x^{15}-x^{16}+1}{(x-1)^2}\end{aligned}$ $ =\frac{15 x^{16}-16 x^{15}+1}{(x-1)^2} $ On comparing LHS $=$ RHS, we get $ p=16 	ext { and } q=15 $ $ 	herefore \quad(p, q)=(16,15) $

If $\frac{d}{d x}\left[(x+1)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\right]$ $=\left(15 x^p-16…

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