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<mdoxml version="1.0"><br></br>
<p>Sum the series $$1 \times 1! + 2 \times 2! + 3 \times 3! +...+n \times n!$$</p>
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Thank you for these solutions to Shahnawaz Abdullah; Daniel, Liceo
Scientifico Copernico, Torino, Italy; Anderthan, Saratoga High
School; Andrei, School 205, Bucharest, Romania; David, Queen Mary's
Grammar School, Walsall; Paddy, Peter, Greshams School, Holt,
Norfolk; Ngoc Tran, Nguyen Truong To High School (Vietnam); Chris,
St. Bees School; Dorothy, Madras College; A Ji and Hyeyoun, St.
Paul's Girls' School; and Yatir from Israel. <br></br>
<br></br>
To prove that $k \times k! = (k+1)! - k!$. <br></br>
<br></br>
If we take $k!$ out as a factor from the right hand side of the
equation, we are left with $k! \times ((k+1)-1)$ which simplifies
to $k \times k!$, as required. <br></br>
<br></br>
Now we sum the series $1 \times 1!+.....n \times n!$ <br></br>
<br></br>
As we have proved, $n \times n!$ is equal to $(n+1)! - n!$ and
therefore $(n-1) \times (n-1)!$ is equal to $(n-1+1)! - (n-1)!$
which simplifies to $n! - (n-1)!$. If we add the two results, we
find that $n!$ cancels. If we sum the series from 1 to $n$, we find
that all of the terms cancel except for $(n+1)!$ and $-(1!)$. Thus
the sum of all numbers of the form $r \times r!$ from $1$ to $n$ is
equal to $(n+1)! - 1$.<br></br></mdoxml>
<mdoxml version="1.0"><br></br>
<span style="font-weight: bold;">Possible support</span> <br></br>
Try the problems <a href="http://nrich.maths.org/public/viewer.php?obj_id=329&part=index">
Natural Sum,</a> <a href="http://nrich.maths.org/public/viewer.php?obj_id=2275&part=index">
More Sequences and Series,</a> and <a href="http://nrich.maths.org/public/viewer.php?obj_id=297&part=index">
OK Now Prove It</a> <br></br>
Read the article <a href="http://nrich.maths.org/public/viewer.php?obj_id=4718&part=index">
Proof by Induction.</a> <br></br>
<br></br>
<span style="font-weight: bold;">Possible extension</span> <br></br>
<a href="http://nrich.maths.org/public/viewer.php?obj_id=267&part=index">
Telescoping Series</a> <br></br></mdoxml>
<mdoxml version="1.0"><p>First prove that $k \times k! = (k+1)! - k!$</p>
<p><br></br>
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Seriesly
Prove that k.k! = (k+1)! - k! and sum the series 1.1! + 2.2! + 3.3!
+...+n.n!
Numbers and the Number System
Factorials
Advanced Algebra
Summation of series
Admin
Short problems