Series sums
Let S1 = 1 , S2 = 2 + 3, S3 = 4 + 5 + 6 ,........ Calculate S17.
Age
14 to 16
Challenge level
Problem
Let
$S_1 = 1$
$S_2 = 2 + 3$
$S_3 = 4 + 5 + 6$, $\cdots$
Calculate $S_{17}$
Student Solutions
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 $S_{17} = 17 \times\frac{17^2+1}{2} = 17 \times{290\over2} = 2465$