Convert Summation into Formula

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 $\dfrac{(n+1)(n+2)}{2} = \dbinom{n+2}{2}$.

You can follow this link for a detailed explanation, but in short

$$ 1 + 2 + \cdots + k = \frac{k(k + 1)}{2} $$

In your case $k = n + 1$

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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

$$ 2S = k (k + 1) $$