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Calculate sum 3!/1!+4!/2!+-----+(n+2)!/n!

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luvgoj | Student, Undergraduate | Honors

Posted July 7, 2012 at 4:18 PM via web

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Calculate sum 3!/1!+4!/2!+-----+(n+2)!/n!

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 7, 2012 at 4:31 PM (Answer #1)

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You should notice that you may write the general term as `((k+2)!)/(k!)`

You need to remember that `((k+2)!) = 1*2*3*....*k*(k+1)*(k+1)`  and `k!= 1*2*3*...*k` , hence, you may write `((k+2)!) ` as `k!*(k+1)*(k+2).` `((k+2)!)/(k!) =(k!*(k+1)*(k+2))/(k!)`

 

Reducing by `k! ` yields:

`((k+2)!)/(k!) = (k+1)*(k+2)`

Hence, you may write the summation such that:

`sum_(k=1)^n((k+2)!)/(k!) =sum_(k=1)^n (k+1)*(k+2)`

You need to open the brackets such that:

`(k+1)*(k+2) = k^2 + 2k + k + 2`

`(k+1)*(k+2) = k^2 + 3k + 2`

Hence `sum_(k=1)^n (k+1)*(k+2) =sum_(k=1)^n (k^2 + 3k + 2)`

`sum_(k=1)^n(k^2 + 3k + 2) =sum_(k=1)^n k^2 +sum_(k=1)^n 3k +sum_(k=1)^n 2`

`sum_(k=1)^n(k^2 + 3k + 2) = (n(n+1)(2n+1))/6 + 3*n(n+1)/2 + 2n`

`sum_(k=1)^n(k^2 + 3k + 2) = (n(n+1)(2n+1))/6 + 9n(n+1)/6 + 12n/6`

`sum_(k=1)^n (k^2 + 3k + 2) = n(n^2 + 3n + 1 + 9n + 9 + 12)/6`

`sum_(k=1)^n (k^2 + 3k + 2) = n(n^2 + 12n + 22)/6`

Hence, evaluating the given sum yields `sum_(k=1)^n ((k+2)!)/(k!) =n(n^2 + 12n + 22)/6.`

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