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# Series Sums

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,

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Age 14 to 16

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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 $S_{17} = 17 \times\frac{17^2+1}{2} = 17 \times{290\over2} = 2465$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.

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