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Telescoping Series

Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.

Degree Ceremony

What does Pythagoras' Theorem tell you about these angles: 90°, (45+x)° and (45-x)° in a triangle?

OK! Now Prove It

Make a conjecture about the sum of the squares of the odd positive integers. Can you prove it?

Seriesly

Age 16 to 18
Challenge Level

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 $(n+1)! - 1$.