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Evaluate sum(r=1)^(n)rxxr!...

Evaluate `sum_(r=1)^(n)rxxr!`

Text Solution

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We have, `underset(r=1)overset(n)(sum)rxxr!=underset(r=1)overset(n)(sum){(r+1)-1}r!=underset(r=1)overset(n)(sum)(r+1)!-r!`
`=(n+1)!-1!`
[put r=n in (r+1)! Annd r=1 is r!]
`=(n+1)!-1`
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