# Question #2e7bb

Mar 31, 2017

Is this question correct?
Should the equation for each term be:

${a}_{i} = \frac{1}{\left(3 i - 1\right) \left(3 i + 2\right)}$

#### Explanation:

Set n = 1

${a}_{n} \to {a}_{1} = \frac{1}{\left(3 \times 1\right) - 1} \times \left(\left(3 \times 1\right) - 2\right) = \frac{1}{2} \times 1 = \frac{1}{2}$

This is different to $\frac{1}{2.5}$
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$\textcolor{b l u e}{\text{The expression that gives each term:}}$

The sequence of values is represented by:

${a}_{i} = \frac{1}{\left[2 + 3 \left(i - 1\right)\right] \textcolor{w h i t e}{.} \left[2 + 3 i\right]} \text{ " = " } \frac{1}{\left(3 i - 1\right) \left(3 i + 2\right)}$

${a}_{1} = \frac{1}{\left(3 - 1\right) \left(3 + 2\right)} \text{ "=" } \frac{1}{2 \times 5}$

${a}_{2} = \frac{1}{\left(6 - 1\right) \left(6 + 2\right)} \text{ "=" } \frac{1}{5 \times 8}$

${a}_{3} = \frac{1}{\left(9 - 1\right) \left(9 + 2\right)} \text{ "=" } \frac{1}{8 \times 11}$
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Mar 31, 2017

$r$ th term of the given series

${t}_{r} = \frac{1}{\left(2 , 5 , 8. . . r \text{th term")(5,8,11...r"th term}\right)}$

$\implies {t}_{r} = \frac{1}{\left(2 + \left(r - 1\right) \times 3\right) \left(5 + \left(r - 1\right) \times 3\right)}$

$\implies {t}_{r} = \frac{1}{\left(3 r - 1\right) \left(3 r + 2\right)}$

$\implies {t}_{r} = \frac{1}{3} \times \frac{3}{\left(3 r - 1\right) \left(3 r + 2\right)}$

$\implies {t}_{r} = \frac{1}{3} \frac{\left(3 r + 2\right) - \left(3 r - 1\right)}{\left(3 r - 1\right) \left(3 r + 2\right)}$

$\implies {t}_{r} = \frac{1}{3} \left(\frac{1}{3 r - 1} - \frac{1}{3 r + 2}\right)$

Putting $r = 1 , 2 , 3. \ldots \text{.upto } n$ and adding we get

For $r = 1 , \text{ "" }$ ${t}_{1} = \frac{1}{3} \left(\frac{1}{2} - \cancel{\frac{1}{5}}\right)$

For $r = 2 , \text{ "" }$ ${t}_{2} = \frac{1}{3} \left(\cancel{\frac{1}{5}} - \cancel{\frac{1}{8}}\right)$

For $r = 3 , \text{ "" }$ ${t}_{3} = \frac{1}{3} \left(\cancel{\frac{1}{8}} - \cancel{\frac{1}{11}}\right)$

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For $r = n - 1 , \text{ }$ ${t}_{n - 1} = \frac{1}{3} \left(\cancel{\frac{1}{3 n - 4}} - \cancel{\frac{1}{3 n - 1}}\right)$

For $r = n , \text{ "" }$${t}_{n} = \frac{1}{3} \left(\cancel{\frac{1}{3 n - 1}} - \frac{1}{3 n + 2}\right)$

$\text{Sum upto n terms } {S}_{n}$

$= \frac{1}{3} \left(\frac{1}{2} - \frac{1}{3 n + 2}\right) = \frac{3 n + 2 - 2}{3 \cdot 2 \cdot \left(3 n + 2\right)} = \frac{n}{2 \cdot \left(3 n + 2\right)}$