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If Sigma(r=1)^(n) Tr=n/8(n+1)(n+2)(n+3) ...

If `Sigma_(r=1)^(n) T_r=n/8(n+1)(n+2)(n+3)` then find `Sigma_(r=1)^(n) 1/T_r`

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The correct Answer is:
`(n(n+3))/(2(n+1)(n+2))`
`T_(n)=sum_(r=1)^(n)T_(r)-sum_(r=1)^(n-1)T_(r)`
`=(n(n+1)(n+2)(n+3))/8-((n-1)n(n+1)(n+2))/8`
`=(n(n+1)(n+2))/2`
`therefore1/(T_(r))=2/(r(r+1)(r+2))=(r+2-r)/(r(r+1)r+2))`
`=1/(r(r+1))-1/((r+1)(r+2)`
`=V(r )-V(r+1)`
`thereforesum_(r=1)^(n)1/(T^(r))=sum_(r=1)^(n)(V(r )-V(r+1))`
`=V(1)-V(n+1)`
`=1/2-1/((n+1)(n+2))`
`=(n(n+3))/(2(n+1)(n+2))`
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