Find the sum `Sigma_(r=1)^(n) 1/(r(r+1)(r+2)(r+3))`
Also,find `Sigma_(r=1)^(oo) 1/(r(r+1)(r+2)(r+3))`

Explanation

`T_(r)=1/(r(r+1)(r+2)(r+3))`
`=(r+3-r)/(3(r(r+1)(r+2)(r+3)))`
`=1/3(1/(r(r+1)(r+2))-1/((r+1)(r+2)(r+3)))`
`1/3(V(r )-V(r+1))`, where V( r)=`1/(r(r+1)(r+2))`
`therefore` Required sum=`sum_(r=1)^(n)1/3(V(r )-V(r+1))`
`=1/3(V(1)-V(n+1))`
`=1/3(1/6-1/((n+1)(n+2)(n+3)))`
`sum_(r=1)^(oo)1/(r(r+1)(r+2)(r+3))`
`=lim_(ntooo)1/3(1/6-1/((n+1)(n+2)(n+3)))=1/3xx1/6=1/18`
Note : If we have general as `(1)/(a_(1)a_(2)a_(3)...a_(n))`, where `a_(1), a_(2), a_(3),....,a_(n)` are in A.P., then we multiply and divided by the difference `a_(n) - a_(1)`.