Weekly Challenge 36: Seriesly

7290 /www/nrich/html/content/id/7290/ 1 2011-05-11T13:28:54 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/315&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/267&part=">Try this next</a> </li> <li> <a href="https://nrich.maths.org/discus/messages/27/27.html">Ask NRICH</a> </li> <li> <a href="http://nrich.maths.org/1437&part=">Read all about it</a> </li> <li> <a href="http://nrich.maths.org/7289&part=solution">Last week's solution</a> </li> </ul> <div> Prove that $$k \times k! = (k+1)! - k!$$ and sum the series $$1 \times 1! + 2 \times 2! + 3 \times 3! +...+n \times n!$$ </div> <div class="framework"> Did you know ... ? A telecoping series is a series that can be written as the difference of two expressions in such a way that almost all the terms cancel with the following or preceding term leaving a few terms which can be combined to give the sum of the series. See the Wikipedia article <a href="http://en.wikipedia.org/wiki/Telescoping_series">Telescoping Series.</a> </div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">This task previously appeared on the main NRICH site, with solutions as follows: 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. To prove that $k \times k! = (k+1)! - k!$. 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. Now we sum the series $1 \times 1!+.....n \times n!$ 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 $(n+1)! - 1$. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">When proving an identity it is usually best to start with the more complicated expression and simplify it so this time start with the expression on the right hand side $(k+1)! - k!$. The algebra is easy. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">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. To prove that $k \times k! = (k+1)! - k!$. 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. Now we sum the series $1 \times 1!+.....n \times n!$ 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 $(n+1)! - 1$. </mdoxml> 2 3 0 0 0 0 1 Weekly Challenge 36: 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 Collections Weekly Challenge