1.

1). 6.322). 63). 44). 4.32

Answer»

$(\eqalign{ & 2\sqrt {16 - 6\sqrt {16 - 6\sqrt {16 - \ldots } } } = k \cr & \therefore\ 2\sqrt {16 - 3 \times 2\sqrt {16 - 3 \times 2\sqrt {16 - 3 \times 2} \sqrt {16 - \ldots } } } = k \cr & \therefore \ 2\sqrt {16 - 3K} = k \cr})$

∴ 4 (16 - 3k) = k2

 ∴ k2 + 12k – 64 = 0

∴ k2 + 16K – 4k – 64 = 0

∴ (k + 16)(k – 4) = 0

k ≠ -16

∴ k = 4



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