Find the sum 3/(1!+2!+3!)+4/(2!+3!+4!...

Find the sum
`3/(1!+2!+3!)+4/(2!+3!+4!)+...+1000/(998!+999!+1000!)`

Text Solution

Verified by Experts

The correct Answer is:

`1/2-1/(1000!)`

`T_(n)=(n+2)/(n!+(n+1)!+(n+2)!)`
`=(n+2)/(n![1+(n+1)+(n+1)(n+2)])`
`=(n+2)/(n!(n^(2)+4n+4))=1/(n!(n+2))`
`=(n+1)/((n+2)!)=((n+2)-1)/((n+2)!)=1/((n+1)!-1/((n+2)!)`
`thereforeS=sum_(n=1)^(998)T_(n)=sum_(n=1)^(998)(1/((n+1)!)-1/((n+2)!))=1/2-1/(1000!)`

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Find the sum `3/(1!+2!+3!)+4/(2!+3!+4!)+...+1000/(998!+999!+1000!)`

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CENGAGE-PROGRESSION AND SERIES-Exercise 5.9

Find the sum
`3/(1!+2!+3!)+4/(2!+3!+4!)+...+1000/(998!+999!+1000!)`