Answer The Following Question ?

#1/(1 . 4)#+#1/(4 . 7)#+#1/(7 . 10)##.........# Summation Of #n^(th)# term ??

2 Answers
George C.
Aug 1, 2018

#sum_(n=1)^N 1/((3n-2)(3n+1)) = N/(3N+1)#

Explanation:

The general term can be written:

#a_n = 1/((3n-2)(3n+1)) = 1/3(1/(3n-2)-1/(3n+1))#

So:

#sum_(n=1)^N a_n = sum_(n=1)^N 1/3(1/(3n-2)-1/(3n+1))#

#color(white)(sum_(n=1)^N a_n) = 1/3(sum_(n=1)^N 1/(3n-2)-sum_(n=1)^N 1/(3n+1))#

#color(white)(sum_(n=1)^N a_n) = 1/3(sum_(n=1)^N 1/(3n-2)-sum_(n=2)^(N+1) 1/(3n-2))#

#color(white)(sum_(n=1)^N a_n) = 1/3(1+color(red)(cancel(color(black)(sum_(n=2)^N 1/(3n-2))))-color(red)(cancel(color(black)(sum_(n=2)^N 1/(3n-2)))) - 1/(3N+1))#

#color(white)(sum_(n=1)^N a_n) = 1/3(1-1/(3N+1))#

#color(white)(sum_(n=1)^N a_n) = N/(3N+1)#

maganbhai P.
Aug 1, 2018

#S_n=n/(3n+1)#

Explanation:

Let ,

#S=1/(1*4)+1/(4*7)+1/(7*10)+...+1/((3n-2)(3n+1))#

#:.t_n=1/((3n-2)(3n+1))=1/3[1/(3n-2)-1/(3n+1)]#

#:.S_n=1/3sum_(r=1)^n [1/(3r-2)-1/(3r+1)]#

#:.S_n=1/3[(1/1-cancel(1/4))color(white)(........................)lArrcolor(red)( r=1#

#color(white)(..............)+(cancel(1/4)-cancel(1/7))color(white)(........................)lArr color(red)(r=2#

#color(white)(..............)+(cancel(1/7)-cancel(1/10))color(white)(........................)lArrcolor(red)( r=3#

#color(white)(..............)+(cancel(1/10)-cancel(1/13))color(white)(........................)lArrcolor(red)( r=4#

#color(white)(..............)+cancel(..).................#

#color(white)(..............)+...................#

#color(white)(..............)+.............cancel(......#

#color(white)(..............)+(cancel(1/(3n-5))-cancel(1/(3n-2)))color(white)(..............)lArrcolor(red)( r=n-1#

#color(white)(..............)+(cancel(1/(3n-2))-1/(3n+1))]color(white)(...............)lArrcolor(red)( r=n#

#:.S_n=1/3[1/1-1/(3n+1)]#

#:.S_n=1/3[(3n+1-1)/(3n+1)]=1/3[(3n)/(3n+1)]#

#:.S_n=n/(3n+1)#

Related questions
Impact of this question
1244 views around the world
You can reuse this answer
Creative Commons License