Question 100

# If $$a_{1}=\frac{1}{2\times5},a_{2}=\frac{1}{5\times8},a_{3}=\frac{1}{8\times11},...,$$ then $$a_{1}+a_{2}+a_{3}+...+a_{100}$$ is

Solution

$$a_{100} = \frac{1}{ (3\times100 -1) \times (3\times100 + 2)}= \frac{1}{ 299 \times 302}$$

$$\frac{1}{2\times5} = \frac{1}{3} \times (\frac{1}{2} - \frac{1}{5})$$

$$\frac{1}{5\times8} = \frac{1}{3} \times (\frac{1}{5} - \frac{1}{8})$$

$$\frac{1}{8\times11} = \frac{1}{3} \times (\frac{1}{8} - \frac{1}{11})$$
....

$$\frac{1}{299\times302} = \frac{1}{3} \times (\frac{1}{299} - \frac{1}{302})$$

Hence $$a_{1}+a_{2}+a_{3}+...+a_{100}$$ = $$\frac{1}{3} \times (\frac{1}{2} - \frac{1}{5})$$ + $$\frac{1}{3} \times (\frac{1}{5} - \frac{1}{8})$$ + $$\frac{1}{3} \times (\frac{1}{8} - \frac{1}{11})$$ + ... + $$\frac{1}{3} \times (\frac{1}{299} - \frac{1}{302})$$

= $$\frac{1}{3} \times (\frac{1}{2} - \frac{1}{302})$$

= $$\frac{25}{151}$$

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