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F given^Pn xB/yC & z given: find F itrepeated r times.
[??]( r^Occurrences of F, always = z. Therefore new value
of z will be r
For yC substitute x(n – 1)C (2)
zF in Pn xB//(x(n – 1)C) becomes rF in P(rn/z) rx/z B//((rx/z)((rn-z)/z)C)
[Example: In P6 3B/15C, a certain F occurs 3 times, and z accordingly = 3.
Required, F repeated twice.
P6 3B/15C = P6 3B/(3(6-1)C)
∴ x = 3, z = 3, n = 6 &r = 2. (=z', new z.)
Knot required is P(2 x 6 / 3) (2 x 3/3) B//((2 x 3/3)((2 x 6) – 3)/3)C
= P4 2B//2(9/3)C
P4 2B//6C], z = 2
(Figure of 8 F.)
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