If P_1, P_2, P_3, ldots, P_n are n points on | vedclass

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(C) Given that $P_1, P_2, \ldots, P_n$ are points on the line $y=x$. Since $O$ is the origin $(0,0)$,the distance $OP_k$ for any point $P_k(x_k, x_k)$

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$7$

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(C) Given that $P_1, P_2, \ldots, P_n$ are points on the line $y=x$. Since $O$ is the origin $(0,0)$,the distance $OP_k$ for any point $P_k(x_k, x_k)$ is given by $OP_k = \sqrt{x_k^2 + x_k^2} = x_k \sqrt{2}$.Given $OP_1 = 1$,we have $x_1 \sqrt{2} = 1$,so $x_1 = \frac{1}{\sqrt{2}}$.The recurrence relation is $OP_n = n(OP_{n-1})$.For $n=2$,$OP_2 = 2(OP_1) = 2(1) = 2$.For $n=3$,$OP_3 = 3(OP_2) = 3(2) = 6$.For $n=4$,$OP_4 = 4(OP_3) = 4(6) = 24$.In general,$OP_n = n 	imes (n-1) 	imes \ldots 	imes 1 = n!$.We are given $P_n = (2520 \sqrt{2}, 2520 \sqrt{2})$.Thus,$OP_n = \sqrt{(2520 \sqrt{2})^2 + (2520 \sqrt{2})^2} = \sqrt{2 	imes (2520^2 	imes 2)} = \sqrt{4 	imes 2520^2} = 2 	imes 2520 = 5040$.So,$n! = 5040$.Since $7! = 7 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1 = 5040$,we have $n=7$.

If $P_1, P_2, P_3, \ldots, P_n$ are $n$ points on the line $y=x$ all lying in the first quadrant,such that $OP_n = n(OP_{n-1})$ ($O$ is origin),$OP_1 = 1$ and $P_n = (2520 \sqrt{2}, 2520 \sqrt{2})$,then $n=$

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