Finding the Sum of an Arithmetic Sequence
Formula
The arithmetic sequence $a_{n} = a+(n-1)d$ with the first term $a$ and the common difference $d$ $$ \sum_{k=1}^{n} a_{k}= {{n \left\{ 2a + (n-1)d \right\} } \over {2}} $$
Explanation
Although this is a series that you might look at once and never write down again in this form, never forget its proof. Even if the proof is simple and straightforward, it’s crucial to write it down by hand at least once to get familiar with it.
The most frequently used sum of an arithmetic sequence is the sum up to natural number $n$. In this case, it becomes an arithmetic sequence where $a=1$ and $d=1$.
Sum of Natural Numbers
$$ \begin{align*} \sum_{k=1}^{n} {k} = {{n(n+1)} \over {2}} \end{align*} $$
If you’re a student preparing for exams, you might find yourself using it so often that you memorize it by heart, like $n=10$ being 55 or $n=100$ being 5050. Even as a university student, you might find it surprisingly useful, so this is a formula you shouldn’t forget.
Proof
If you say $$ \begin{align*} S:= \sum_{k=1}^{n} a_{k} \end{align*} $$ Then $$ S= a + (a+d) + \cdots + (a + (n-2)d) + (a + (n-1)d) $$ But if you write this sequence in reverse order $$ S = \left\{ a + (n-1)d \right\} + \left\{ a + (n-2)d \right\} + \cdots + (a+d) + a $$ It’s also true. If you add both sides $$ 2S = \left[ \left\{ 2a + (n-1)d \right\} + \left\{ 2a + (n-1)d \right\} + \cdots +\left\{ 2a + (n-1)d \right\} + \left\{ 2a + (n-1)d \right\} \right] $$ Since you’ve added a total of $n$ terms $$ 2S = n \left\{ 2a + (n-1)d \right\} $$ Dividing both sides by 2 leads to $$ \sum_{k=1}^{n} a_{k}= {{n \left\{ 2a + (n-1)d \right\} } \over {2}} $$
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