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