1, 1, 4, 8, 9, 27, 16, ?

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256

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Solution

The correct option is B 64
The series consists of squares and cubes of consecutive natural numbers i.e. $1^{2}, 1^{3}, 2^{2}, 2^{3}, 3^{2}, 3^{3}, 4^{2}, . . . . .$
So, missing term = $4^{3} = 64$ .

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Similar questions

1, 1, 1, 2, 4, 8, 3, 9, 27, 4, 16, 32

1, 1, 4, 8, 9, 27, 16, ?

{lim}_{n \to \infty} \frac{1 + 1 + 4 + 8 + 9 + 27 + 16 + 64 + . . .}{(1 + 2 + 3 + 4 + 5 + . . .)^{2}} =

3^{1 / 4} {.9}^{1 / 8} {.27}^{1 / 16} {.81}^{1 / 32} . . . . . . . . . .,

\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . = x

\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + . . . = y

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