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Natural Sum

The picture illustrates the sum 1 + 2 + 3 + 4 = (4 x 5)/2. Prove the general formula for the sum of the first n natural numbers and the formula for the sum of the cubes of the first n natural numbers.

Mystic Rose

Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes.

Slick Summing

Watch the video to see how Charlie works out the sum. Can you adapt his method?

Series Sums

Age 14 to 16 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 $S_{17} = 17 \times\frac{17^2+1}{2} = 17 \times{290\over2} = 2465$