Question

Consider an+1 = 1/1+1/an for n = 1,2,.......2008,2009. where a1 = 1. Find the value of a1a2 + a2a3 + a3a4+ .......+ a2008a2009.

• A. 2009/1000
• B. 2009/2008
• C. 2008/2009
• D. 6000/2009
• E. 2008/6000

Answer 1

The Correct Option is C

To solve the given problem, we are provided with a recurrence relation:

\[ a_{n+1} = \frac{1}{1 + \frac{1}{a_n}} = \frac{a_n}{a_n + 1} \]

where \( a_1 = 1 \). We need to find the value of \( a_1a_2 + a_2a_3 + \ldots + a_{2008}a_{2009} \).

First, let's analyze the recurrence relation:

• Given \( a_1 = 1 \), using the recurrence relation, \( a_2 = \frac{1}{1 + \frac{1}{a_1}} = \frac{1}{2} \).
• Similarly, \( a_3 = \frac{a_2}{a_2 + 1} = \frac{\frac{1}{2}}{\frac{1}{2} + 1} = \frac{1}{3} \).

We can observe a potential pattern: \(a_n = \frac{1}{n}\) by mathematical induction for all \(n\).

Proof by Induction:

• Base Case: \( a_1 = 1 \), which matches \( \frac{1}{1} \).
• Inductive Step: Assume \( a_k = \frac{1}{k} \), then \( a_{k+1} = \frac{a_k}{a_k + 1} = \frac{\frac{1}{k}}{\frac{1}{k} + 1} = \frac{1}{k+1} \), hence true for \( k+1 \).

By mathematical induction, the hypothesis holds for all \(n\).

To find the sum \( a_1a_2 + a_2a_3 + \ldots + a_{2008}a_{2009} \), consider:

\[ a_1a_2 + a_2a_3 + \ldots + a_{2008}a_{2009} = \sum_{n=1}^{2008} a_n a_{n+1} = \sum_{n=1}^{2008} \frac{1}{n(n+1)} \]

The sequence \(\frac{1}{n(n+1)}\) can be simplified using partial fraction decomposition:

\[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \]

Therefore, the sum becomes a telescoping series:

\[ \sum_{n=1}^{2008} \left( \frac{1}{n} - \frac{1}{n+1} \right) \]

The telescoping series results in:

\[ \left( 1 - \frac{1}{2009} \right) = \frac{2008}{2009} \]

Thus, the value of \(a_1a_2 + a_2a_3 + \cdots + a_{2008}a_{2009}\) is:

\[ \frac{2008}{2009} \]

Therefore, the correct answer is \(\frac{2008}{2009}\).

Answer 2

The recurrence is: \[ a_{n+1} = \frac{a_n}{a_n + 1}. \] Hence, \[ a_n a_{n+1} = \frac{a_n^2}{a_n + 1}. \]

Notice that: \[ a_n - a_{n+1} = \frac{a_n}{a_n+1} \cdot \frac{a_n}{1} = \frac{a_n^2}{a_n+1} = a_n a_{n+1}. \] Therefore, \[ a_n a_{n+1} = a_n - a_{n+1}. \] 

Thus, the sum telescopes: \[ S = (a_1 - a_2) + (a_2 - a_3) + \cdots + (a_{2008} - a_{2009}). \] Simplifying: \[ S = a_1 - a_{2009}. \]

Since \(a_1 = 1\), \[ S = 1 - a_{2009}. \]

To evaluate \(a_{2009}\), observe the pattern by defining \(b_n = \frac{1}{a_n}\). Then, \[ a_{n+1} = \frac{a_n}{a_n+1} \quad \Rightarrow \quad \frac{1}{a_{n+1}} = \frac{a_n+1}{a_n} = 1 + \frac{1}{a_n}. \] So, \[ b_{n+1} = b_n + 1, \quad \text{with } b_1 = \frac{1}{a_1} = 1. \] Hence, \[ b_n = n \quad \Rightarrow \quad a_n = \frac{1}{n}. \]

Therefore, \[ a_{2009} = \frac{1}{2009}. \] Substituting back: \[ S = 1 - \frac{1}{2009} = \frac{2008}{2009}. \]

Final Answer:
\[ \boxed{\tfrac{2008}{2009}} \]

Options Check:

Option	Value	Status
\(\tfrac{2009}{1000}\)	≈ 2.009	Incorrect
\(\tfrac{2009}{2008}\)	≈ 1.0005	Incorrect
\(\tfrac{2008}{2009}\)	≈ 0.9995	Correct
\(\tfrac{6000}{2009}\)	≈ 2.987	Incorrect
\(\tfrac{2008}{6000}\)	≈ 0.334	Incorrect