Question

Let S₁, S₂, … be the squares such that for each n ≥ 1, the length of the diagonal of S_n is equal to the length of the side of S_n+1. If the length of the side of S₃ is 4 cm, what is the length of the side of S_n?

• A. $\dfrac{2^{n+1}}{2^{1/2}}$
• B. $2 \cdot (n-1)$
• C. $2^{n-1}$
• D. $2^{\frac{n+1}{2}}$
• E. None of these

Hint:

Whenever diagonal-to-side relations appear in square sequences, look for $\sqrt{2}$ ratio, which creates a geometric progression.

Answer 1

The Correct Option is D

Step 1: Relationship between sides of consecutive squares.
For each $n$, \[ \text{side}(S_{n+1}) = \text{diagonal}(S_n) \] If the side of $S_n$ is $a$, then the diagonal is $a\sqrt{2}$. So: \[ \text{side}(S_{n+1}) = \sqrt{2}\cdot \text{side}(S_n) \] Step 2: Identify sequence type.
The sequence of sides is a geometric progression (G.P.) with common ratio $r=\sqrt{2}$. 
Step 3: Express side of $S_n$.
General term: \[ \text{side}(S_n) = a \cdot (\sqrt{2})^{n-1} \] where $a$ is the first term. 
Step 4: Use given value at $S_3$.
\[ \text{side}(S_3) = a \cdot (\sqrt{2})^{2} = a \cdot 2 \] Given side$(S_3)=4$, hence $a\cdot2=4 \Rightarrow a=2$. 
Step 5: Final formula.
So: \[ \text{side}(S_n) = 2 \cdot (\sqrt{2})^{n-1} \] \[ = 2^{1}\cdot 2^{\tfrac{n-1}{2}}=2^{\tfrac{n+1}{2}} \] \[ \boxed{2^{\tfrac{n+1}{2}}} \]