AB, BA, ABC, CBA, ABCD, ?

Question

Let \[ S=\frac{1}{2!5!}+\frac{1}{3!2!3!}+\frac{1}{5!2!1!}+\cdots \text{ up to 13 terms}. \] If $13S=\dfrac{2^k}{n!}$, $k\in\mathbb{N}$, then $n+k$ is equal to

Answer 1

The sequence follows this pattern: Term, Reverse(Term), Term+NextLetter, Reverse(Term+NextLetter), and so on. Example: 1: AB, 2: BA (reverse of 1), 3: ABC (AB + C), 4: CBA (reverse of 3), 5: ABCD, 6: Expected to be the reverse of 5 (ABCD) which is DCBA. This matches option (3).

AB, BA, ABC, CBA, ABCD, ?

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The Correct Option is C

Solution and Explanation

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