Question 65

Consider $$a_{n+1} =\frac{1}{1+\frac{1}{a_{n}}}$$ for $$n = 1,2, ....., 2008, 2009$$ where $$a_{1} = 1$$. Find the value of $$a_{1}a_{2} + a_{2}a_{3} + a_{3}a_{4} + ... + a_{2008}a_{2009}$$.

Solution

Given that $$a_1=1$$ & $$a_{n+1} =\frac{1}{1+\frac{1}{a_{n}}}$$

$$a_2=\ \frac{\ 1}{1+\ \frac{\ 1}{1}}=\ \ \frac{1}{2},\ a_3=\ \frac{\ 1}{1+\ \frac{\ 1}{\left(\frac{1}{2}\right)}}=\frac{1}{3},....$$

This implies, $$\ a_n=\ \frac{1}{n}$$.

Required value$$=a_1a_2+a_2a_3+........+a_{2008}a_{2009}$$

= $$\frac{1}{1}\times\ \frac{1}{2}+\frac{1}{2}\times\ \frac{1}{3}+.....+\frac{1}{2008}\times\ \frac{1}{2009}$$

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

= $$1-\ \frac{1}{2009}$$ 

= $$\frac{2008}{2009}$$

The answer is option C.

Video Solution

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