The inductive step of an inductive proof shows that for k > 0, if E 2' = 2k+l – 1 then E 2 = 2*+2 j=0 In which step of the proof is the inductive hypothesis used? E 2 3(2*+1 – 1) + 2*+1 (Step 2) E, 2 + 2*+1 (Step 1) j=0 %3D = 2· 2k+1 – 1 (Step 3) = 2k+2 - 1 (Step 4) O Step 3 O Step 2 O Step 1 O Step 4

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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The inductive step of an inductive proof shows that for \( k \geq 0 \), if \( \sum_{j=0}^{k} 2^j = 2^{k+1} - 1 \) then \( \sum_{j=0}^{k+1} 2^j = 2^{k+2} - 1 \).

In which step of the proof is the inductive hypothesis used?

\[
\begin{align*}
\sum_{j=0}^{k+1} 2^j & = \sum_{j=0}^{k} 2^j + 2^{k+1} \quad \text{(Step 1)} \\
& = \left(2^{k+1} - 1\right) + 2^{k+1} \quad \text{(Step 2)} \\
& = 2 \cdot 2^{k+1} - 1 \quad \text{(Step 3)} \\
& = 2^{k+2} - 1 \quad \text{(Step 4)}
\end{align*}
\]

Choose one:
- Step 3
- Step 2
- Step 1
- Step 4
Transcribed Image Text:The inductive step of an inductive proof shows that for \( k \geq 0 \), if \( \sum_{j=0}^{k} 2^j = 2^{k+1} - 1 \) then \( \sum_{j=0}^{k+1} 2^j = 2^{k+2} - 1 \). In which step of the proof is the inductive hypothesis used? \[ \begin{align*} \sum_{j=0}^{k+1} 2^j & = \sum_{j=0}^{k} 2^j + 2^{k+1} \quad \text{(Step 1)} \\ & = \left(2^{k+1} - 1\right) + 2^{k+1} \quad \text{(Step 2)} \\ & = 2 \cdot 2^{k+1} - 1 \quad \text{(Step 3)} \\ & = 2^{k+2} - 1 \quad \text{(Step 4)} \end{align*} \] Choose one: - Step 3 - Step 2 - Step 1 - Step 4
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