### Sixational

The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by 6.

### Squaresearch

Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares?

### Loopy

Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?

# Series Sums

##### Stage: 4 Challenge Level:

As each sum develops it should become clear that the last number in each sum is triangular. So for $S_n$, the last number in the sum is the $n^{th}$ triangular number $= n(n + 1)/2$. Bearing this in mind and the fact that the first number in the sum is the $(n - 1)^{th}$ triangular number plus $1$, then,
\begin{eqnarray}S_n &=& \frac{n(n - 1)} {2} + 1 + \frac{n(n - 1)} {2} + 2 + \frac{n(n - 1)} {2} + 3 + \frac{n(n - 1)} {2} + \cdots + \frac{n(n - 1)} {2} + n \\ \; &=& \frac{n^2(n - 1)}{2} + (1 + 2 + 3 + 4 + \cdots + n) \\ \; &=& \frac{n^2(n - 1)}{2} + \frac{n(n + 1)}{2} \\ \; &=& \frac{n(n^2 - n)}{2} + \frac{n(n + 1)}{2} \\ S_n &=& \frac{n(n^2 + 1)}{2}\end{eqnarray}
Therefore $S_{17} = 17 \times\frac{17^2+1}{2} = 17 \times{290\over2} = 2465$