Prove $\sum_{k=p}^{q} c = (q-p+1)c$ where $p$,$q$ are integers, $p$ less than or equal to $q$ .

Question

Before I can prove it, can somebody help me explain the summation formula. Some hints, anything please.

The equation to prove is $$\sum_{k=p}^{q} c = (q-p+1)c$$, where $p$,$q$ are integers, $p$ less than or equal to $q$ .

Answer 1

Let $c = 1$ for convenience. Then the summation asks to count the number of times you add 1. Suppose we added 1 once for each $k = 1, k = 2, ..., k = q$. Then we add 1 $q$ times. But since we start adding at $k = p$, we have to subtract all the 1s we counted in $k = 1, k = 2, ..., k = (p-1)$. Then we have added 1 $(p-1)$ times in the second case. So we must subtract $(p-1)$ from $q$ to get $q-(p-1)=q-p+1$.

Answer 2

Since you are summing over the constant value c (each term in the series is c), you can simply count the number of terms in the sum and multiply this value by c. This method is motivated by the interpretation of multiplication as repeated addition. You must be careful when counting the terms, however, since subtracting the last and first index does not yield the desired result. You can demonstrate this by considering the case when q=p. Here the sum simplifies to c, so there is one term in this series despite the fact that p-q=0.