Discussion

Explanation:

Given, a_{n + 1} = $\frac{1}{1+{\displaystyle \frac{1}{a_n}}}$

⇒ a₁ = 1

⇒ a₂ =  $\frac{1}{1+{\displaystyle \frac{1}{a_1}}}$ = $\frac{1}{2}$

⇒ a₃ =  $\frac{1}{1+{\displaystyle \frac{1}{a_2}}}$ = $\frac{1}{3}$
...
⇒ a_n =  $\frac{1}{n}$

Now: a₁ a₂ + a₂ a₃ + a₃ a₄ + ... + a₂₀₀₈ a₂₀₀₉  = $\frac{1}{1\times 2}$ + $\frac{1}{2\times 3}$ + $\frac{1}{3\times 4}$ + ... + $\frac{1}{2008\times 2009}$

= $\left(\frac{1}{1}-\frac{1}{2}\right)$ + $\left(\frac{1}{2}-\frac{1}{3}\right)$ + $\left(\frac{1}{3}-\frac{1}{4}\right)$ + ... + $\left(\frac{1}{2008}-\frac{1}{2009}\right)$

= $\frac{1}{1}-\frac{1}{2009}$ = $\frac{2008}{2009}$

Hence, option (c).

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