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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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Verified Answer
The correct answer is:
$(16,15)$
$\begin{aligned} & \frac{d}{d x}\left[(x+1)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\right] \\ &=\frac{\left(15 x^\rho-16 x^9+1\right)}{(x-1)^2} \\ & \text { LHS }= \frac{d}{d x}\left[\frac{\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)}{(x-1)}\right] \\ &=\frac{d}{d x}\left[\frac{\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)}{(x-1)}\right] \\ &=\frac{d}{d x}\left[\frac{\left(x^8-1\right)\left(x^8+1\right)}{(x-1)}\right] \\ &=\frac{d}{d x}\left[\frac{\left(x^{16}-1\right)}{(x-1)}\right] \\ &= \frac{(x-1)\left(16 x^{15}\right)-\left(x^{16}-1\right)}{(x-1)^2} \\ &= \frac{16 x^{16}-16 x^{15}-x^{16}+1}{(x-1)^2}\end{aligned}$
$$
=\frac{15 x^{16}-16 x^{15}+1}{(x-1)^2}
$$
On comparing LHS $=$ RHS, we get
$$
p=16 \text { and } q=15
$$
$$
\therefore \quad(p, q)=(16,15)
$$
$$
=\frac{15 x^{16}-16 x^{15}+1}{(x-1)^2}
$$
On comparing LHS $=$ RHS, we get
$$
p=16 \text { and } q=15
$$
$$
\therefore \quad(p, q)=(16,15)
$$
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