Question

Prove by induction the identity \((n 1) \cdot (n 2) \ldots (2n-1) \cdot 2n 1 \cdot 3 \cdot 5 \ldots (2n-1) = 2n\) for any natural \(n\).

A) True
B) False
C) Cannot be determined
D) Insufficient information

Answer 1

B) False

The best answer for the option is B) False

Step-by-step explanation:

The given identity is:

`\((n-1) \cdot (n-2) \cdot \ldots \cdot 2 \cdot 1 \cdot (2n-1) \cdot 2n = 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1) \cdot 2n\)`

To prove this identity by induction:

Base Case (n = 1):

For
`\(n = 1\)`, the left-hand side (LHS) is
`\(1 \cdot 2 = 2\)` and the right-hand side (RHS) is
`\(1 \cdot 2 = 2\)`. This validates the identity for the base case.

Inductive Hypothesis:

Assume the identity holds for
`\(n = k\)`, where:

`\((k-1) \cdot (k-2) \cdot \ldots \cdot 2 \cdot 1 \cdot (2k-1) \cdot 2k = 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2k-1) \cdot 2k\)`

Inductive Step (n = k + 1):

Now, we aim to prove the identity for
`\(n = k + 1\)`. Substituting
`\(n = k + 1\)` into the equation, the LHS becomes:

`\((k) \cdot (k-1) \cdot \ldots \cdot 2 \cdot 1 \cdot (2k+1) \cdot 2(k+1)\)`

And the RHS becomes:

`\(1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2k+1) \cdot 2(k+1)\)`

Upon expansion, the LHS and RHS do not equate. This discrepancy indicates that the identity does not hold true for
`\(n = k + 1\)` based on the assumption of
`\(n = k\)` in the inductive step. Hence, the identity does not hold beyond the base case
`\(n = 1\)` and is false for all natural numbers
`\(n\)`.

The best answer for the option is B) False