Stage: 5 Short Challenge Level:
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
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 $(n+1)! - 1$.