0

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

MJD
  • 65,394
  • 39
  • 298
  • 580
Reza Mohammady
  • 49

3 Answers3

4

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 $a=0$).

Marc van Leeuwen
  • 115,048
  • It should still be fine for $n<m-1$: $\sum_{i=m}^n a = -\sum_{j=n+1}^{m-1} a$ – Andrew Dudzik Sep 06 '14 at 09:44
  • 1
    @you-sir-33433: No, the usual convention for summations is that if the summation range contains no integers, then the sum is $0$. It is true that for integrals one usually defines the integral in the reverse direction to be minus the corresponding forward integral, and something similar can be done for simple summations; however one usually takes the set of valid indices as basis for the definition, rather than counting indices that are excluded by both bound conditions with weight $-1$. I think the latter convention would become rather difficult to maintain for multiple summations. – Marc van Leeuwen Sep 06 '14 at 10:45
  • Strange, I have never heard of this convention. I don't really see any advantages to handling sums differently from integrals, especially when it breaks so many formulas and basic algebraic properties (e.g. $\sum_{a}^b + \sum_{b+1}^c = \sum_a^c$). – Andrew Dudzik Sep 06 '14 at 17:49
0

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 $a$.
$\sum_{i=m}^na=a+a(n-m)=a(n-m+1)$ $[n\ge m-1]$

creative
  • 3,539
0

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

EDIT:

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

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

Adi Dani
  • 16,949
  • 3
    Is there any particular reason to disguise $1$ as $\frac ii$? Especially since $m\leq0\leq n$ is allowed, so $i$ could be $0$. – Marc van Leeuwen Sep 06 '14 at 09:29
  • I find it strangely captivating. A joke maybe? – Andrew Dudzik Sep 06 '14 at 09:45
  • Since $i$ is muted index $\sum_{i=m}^{n}1=\sum_{j=m}^{n}1$ we can write something like that. How you would explain why $\sum_{i=m}^{n}1=n-m+1$ – Adi Dani Sep 06 '14 at 11:24
  • 1
    @AdiDani Because $n-m+1$ is exactly what we write when we mean "add $1$ to itself $n-m+1$ times"? How would you explain why $\frac{m}{m} + \frac{m+1}{m+1} + \cdots + \frac{n}{n} = n-m+1$? – Andrew Dudzik Sep 06 '14 at 17:50
  • $\frac{m}{m}+\cdots+\frac{n}{n}$ is farther away from the answer $n-m+1$ than the original expression $1+\cdots+1$ is. There is literally no point to it, it serves only to distract and mislead. This is why I downvoted this answer. – anon Sep 09 '14 at 02:02