TITLE: Convert Summation into Formula
QUESTION [7 upvotes]: I have the formula for finding the average time of an linear search algorithm as below
\begin{equation*}
AT\ =\frac{\ \sum ^{n+1}_{i\ =\ 1} \ \theta ( i)}{n+1}.
\end{equation*}
The above formula was simplified to become the one below, I am not sure how it was moved from that to this. Any help?
\begin{equation*}
AT\ =\ \frac{\theta (( n\ +\ 1) \ \times \ ( n\ +\ 2) \div 2)}{n\ +\ 1}.
\end{equation*}

REPLY [13 votes]: A linear operator satisfies the following property:
$$\theta(a)+\theta(b) = \theta(a+b)$$
This means that:
$$\sum_{i=1}^{n+1}\theta(i) = \theta\left( \sum_{i=1}^{n+1} i \right)$$
caverac answered about that summation yielding $\dfrac{(n+1)(n+2)}{2} = \dbinom{n+2}{2}$.

REPLY [10 votes]: You can follow this link for a detailed explanation, but in short
$$
1 + 2 + \cdots + k = \frac{k(k + 1)}{2}
$$
In your case $k = n + 1$

EIDT
Proof:
Call $S = 1 + 2 + \cdots + (k - 1) + k $, note that you can also write it in reverse order $S = k + (k - 1) + \cdots + 2 + 1 $, so that
\begin{array}{}
S &=& 1 &+& 2 &+\cdots+& (k-1) &+& k\\
S &=& k &+& (k-1) &+\cdots+& 2 &+& 1\\ \hline
2S &=& (k+1) &+& (k+1) &+\cdots+& (k+1) &+& (k+1)
\end{array}
So you have 
$$
2S = k (k + 1)
$$