Vector Summation Trick (Inspired by Gauss) #
In this article we’re going to be dealing with a little vector sum.
We’re given:
\[ \sum_{i=1}^{100} \begin{bmatrix} i - 1 \\ i \\ i + 1 \end{bmatrix} \]And if we don’t recognize this sum immediately, it’s usually a good idea to try a few terms and look for a pattern.
Trying a Few Terms #
When \( i = 1 \):
\[ \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \]When \( i = 2 \):
\[ \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \]When \( i = 3 \):
\[ \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} \]… and so on, until \( i = 100 \):
\[ \begin{bmatrix} 99 \\ 100 \\ 101 \end{bmatrix} \]So, what are we really summing here?
Adding Vectors #
When you add column vectors, you add component-wise. So this becomes:
[ \sum_{i=1}^{100} \begin{bmatrix} i - 1 \ i \ i + 1 \end{bmatrix} #
\begin{bmatrix} \sum_{i=1}^{100} (i - 1) \ \sum_{i=1}^{100} i \ \sum_{i=1}^{100} (i + 1) \end{bmatrix} ]
Now we can break each of these down.
Middle Term: Gauss Trick #
Everyone knows how to add:
\[ 1 + 2 + 3 + \dots + 100 \]And for that, there’s a trick supposedly discovered by Gauss in elementary school.
Pair numbers from the ends:
- \( 1 + 100 = 101 \)
- \( 2 + 99 = 101 \)
- \( 3 + 98 = 101 \)
- …
There are 50 such pairs:
\[ \sum_{i=1}^{100} i = 50 \times 101 = 5050 \]So the middle element in our vector sum is:
\[ \sum_{i=1}^{100} i = 5050 \]Top Term: Each \( i - 1 \) #
This is just shifting each term down by 1, so it’s:
\[ \sum_{i=1}^{100} (i - 1) = \sum_{i=1}^{100} i - 100 = 5050 - 100 = 4950 \]Bottom Term: Each \( i + 1 \) #
Each term is bumped up by 1:
\[ \sum_{i=1}^{100} (i + 1) = \sum_{i=1}^{100} i + 100 = 5050 + 100 = 5150 \]Final Answer #
So, the entire sum simplifies to:
\[ \begin{bmatrix} 4950 \\ 5050 \\ 5150 \end{bmatrix} \]Credits #
All explanations and problems written by Adrian Hernandez Vega, unless otherwise noted.
Prefer video? Watch the full explanation here: