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2011-02-01T00:00:01
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<mdoxml version="1.0"><div>Let $S_1 = 1$, $S_2 = 2 + 3$, $S_3 = 4 + 5 + 6$,
$\cdots$</div>
<br></br>
<div>Calculate $S_{17}$</div>
<br></br></mdoxml>
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<mdoxml version="1.0"><br></br>
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,
<div class="math">\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}</div>
Therefore $S_{17} = 17 \times\frac{17^2+1}{2} = 17 \times{290\over2} = 2465$<br></br></mdoxml>
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Series Sums
Let S1 = 1 , S2 = 2 + 3, S3 = 4 + 5 + 6 ,........ Calculate S17.
Sequences, Functions and Graphs
Sequences
Sequences, Functions and Graphs
Arithmetic sequence