Find the odd number from the given sequence: | vedclass

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Question

(C) The sequence consists of two alternating patterns.The numbers at odd positions $(1^{st}, 3^{rd}, 5^{th}, 7^{th})$ are squares of odd numbers: $1 =

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) The sequence consists of two alternating patterns.The numbers at odd positions $(1^{st}, 3^{rd}, 5^{th}, 7^{th})$ are squares of odd numbers: $1 = 1^2$,$9 = 3^2$,$25 = 5^2$,$49 = 7^2$.The numbers at even positions $(2^{nd}, 4^{th}, 6^{th})$ are cubes of even numbers: $16 = 2^4$ (or $4^2$),$64 = 4^3$,$216 = 6^3$.Re-evaluating the pattern: The sequence is $1^2, 2^4, 3^2, 4^3, 5^2, 6^3, 7^2$. Wait,let's look at the powers: $1^2, 4^2, 3^2, 8^2, 5^2, 216, 7^2$. This is not consistent.Let's check the powers again: $1^2, 2^4, 3^2, 4^3, 5^2, 6^3, 7^2$. Actually,the sequence is $1^2, 2^4, 3^2, 4^3, 5^2, 6^3, 7^2$. The term $216$ is $6^3$. If the pattern is $n^2$ for odd positions and $n^3$ for even positions,then $16$ should be $2^3 = 8$. Since $16$ is $4^2$,the sequence is $1^2, 4^2, 3^2, 8^2, 5^2, dots$ which is not correct.Correct pattern: Odd positions are $1^2, 3^2, 5^2, 7^2$. Even positions are $2^4, 4^3, 6^3$. This is inconsistent. Most likely pattern: $1^2, 2^4, 3^2, 4^3, 5^2, 6^3, 7^2$. The odd positions are $1^2, 3^2, 5^2, 7^2$. The even positions are $2^4, 4^3, 6^3$. The term $216$ is $6^3$. The term $64$ is $4^3$. The term $16$ is $2^4$. The odd one is $16$ because it is $2^4$ while others are $n^3$.

Find the odd number from the given sequence: $1, 16, 9, 64, 25, 216, 49$.

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