Is it correct that $ \sum_{i=m}^na=(nm+1)a$

Question

I study this in a book. Is it correct? Why? $$ \sum_{i=m}^na=(nm+1)a$$

Answer 1

Of course this is wrong (try $n=m$). The factor in parentheses should be $n-m+1$. And even then this is only valid if $n\geq m-1$ (or $a=0$).

Answer 2

The result is incorrect. $\sum_{i=m}^na=(nm+1)a$ for $n=m$ results in $(m^2+1)a$ but it should actually give you $a$.
$\sum_{i=m}^na=a+a(n-m)=a(n-m+1)$ $[n\ge m-1]$

Answer 3

$$\sum_{i=m}^na=a\sum_{i=m}^n1=a\sum_{i=m}^n\frac{i}{i}=a\left(\frac{m}{m}+\frac{m+1}{m+1}+\cdots+\frac{n}{n}\right)=a(n-m+1)$$

EDIT:

From definition $$\sum_{i=m}^na_i=a_m+a_{m+1}+...+a_n$$

for $a_n=...=a_m=a$ follow that $$\sum_{i=m}^na=a(n-m+1)$$