Logic

Pattern-1,1,1,2,4,8,3,9,?,4,16

${ 1 }^{ 1 },{ 1 }^{ 2 },1^{ 3 },{ 2 }^{ 1 },{ 2 }^{ 2 },{ 2 }^{ 3 },{ 3 }^{ 1 },{ 3 }^{ 2 },?,{ 4 }^{ 1 },{ 4 }^{ 2 }$

${ 1 }^{ 1 },{ 1 }^{ 2 },1^{ 3 },{ 2 }^{ 1 },{ 2 }^{ 2 },{ 2 }^{ 3 },{ 3 }^{ 1 },{ 3 }^{ 2 },{ 3 }^{ 3 },{ 4 }^{ 1 },{ 4 }^{ 2 }$

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

Answer- $\boxed{27}$

The pattern follows like this:

Let $T_r$ denote the $r$ th term of the sequence, then:

If $r=3k$ for some $k \in Z^+$ , $T_r = k^3$

If $r=3k+1$ for some $k \in Z^+$ , $T_r = k+1$

If $r=3k+2$ for some $k \in Z^+$ , $T_r = (k+1)^2$

Now, just put the value of $k$ randomly (according to desired $r$ ) to get any term of the sequence quickly. Here, required $r = 9 = 3 \times 3$ .

So, $k=3$

Hence, $T_9 = 3^3$ $=\boxed{27}$

As the pattern is 1,1,1,2,4,8,3,9,?,4,16 We can see that at first the number is there then its square and after that its cube 1 is the number, its square is 1 and its cube is 1. 2 is the number , its square is 4 and its cube is 8. So, 3 is the number, 9 is the square and 27 is its cube. therefore the answer is 27.