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Answer 1 · 2017-03-31T09:01:27

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

$a_i=1/((3i-1)(3i+2))$

Set n = 1

$a_n->a_1=1/((3xx1)-1)xx ((3xx1)-2) =1/2xx1=1/2$

This is different to $1/(2.5)$
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$color(blue)("The expression that gives each term:")$

The sequence of values is represented by:

$a_i=1/([2+3(i-1)]color(white)(.)[2+3i])" " = " "1/((3i-1)(3i+2))$

$a_1=1/((3-1)(3+2))" "=" "1/(2xx5)$

$a_2=1/((6-1)(6+2))" "=" "1/(5xx8)$

$a_3=1/((9-1)(9+2))" "=" "1/(8xx11)$
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Answer 2 · 2017-03-31T15:28:40

$r$ th term of the given series

$t_r=1/((2,5,8...r"th term")(5,8,11...r"th term"))$

$=>t_r=1/((2+(r-1)xx3)(5+(r-1)xx3))$

$=>t_r=1/((3r-1)(3r+2))$

$=>t_r=1/3xx3/((3r-1)(3r+2))$

$=>t_r=1/3((3r+2)-(3r-1))/((3r-1)(3r+2))$

$=>t_r=1/3(1/(3r-1)-1/(3r+2))$

Putting $r=1,2,3....".upto " n$ and adding we get

For $r=1," "" "$ $t_1=1/3(1/2-cancel(1/5))$

For $r=2," "" "$ $t_2=1/3(cancel(1/5)-cancel(1/8))$

For $r=3," "" "$ $t_3=1/3(cancel(1/8)-cancel(1/11))$

$---------$

For $r=n-1," "$ $t_(n-1)=1/3(cancel(1/(3n-4))-cancel(1/(3n-1)))$

For $r=n," "" "$ $t_n=1/3(cancel(1/(3n-1))-1/(3n+2))$

$"Sum upto n terms "S_n$

$=1/3(1/2-1/(3n+2))=(3n+2-2)/(3*2*(3n+2))=n/(2*(3n+2))$