Question

Question #2e7bb

Answer 1

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

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

Explanation:

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

`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))`