Question

Hello everyone, I need help with an exercise about summations, why p + (p + 1) + ... + (q - 1) + q can be resolved bye the equation $\frac{(q+p)(q-p+1)}{2}$.
I just want to know why the summation can be resolved by this equation.

Answer 1

Let $S$ denote the sum in question:

$S=p+(p+1)+\cdots+(q-1)+q$

Reorder the terms as

$S=q+(q-1)+\cdots+(p+1)+p$

Note that each corresponding term in the sums add to $p+q$:

$p+q=p+q$
$(p+1)+(q-1)=p+q$
and so on.

So adding both sums together gives

$2S=(p+q)+(p+q)+\cdots+(p+q)+(p+q)$

There are $p-q+1$ instances of $p+q$. ($p-q$ is the difference between the first and last terms of the sum, i.e. the number of terms after $p$ to count up to $q$. Adding 1 will include $p$ in the count.) So,

$2S=(p-q+1)(p+q)$
$\implies S=\dfrac{(q+p)(q-p+1)}2$