## Loser66 one year ago We have $\sum_{j =1}^n j = \dfrac{n(n+1)}{2}$ How about the limit from m to n? I meant $\sum_{j =m}^n j=??$ Please, help

1. ParthKohli

$= \sum_{1}^{n}j - \sum_{1}^{m-1}j$

2. ParthKohli

sorry for lack of notation

3. imqwerty

terms from m to n = n-m+1 so summation wuld be- $\frac{ (n-m+1)(m+n) }{ 2 }$

4. imqwerty

number of terms =n-m+1 1st term =m and common diff=1 so just put this in the formula of summation of ap :)

5. Loser66

Got it. thank you.

6. imqwerty

np :)

7. Loser66

In general, if we see the lower limit is not 1, when we see n, we replace by what? hehehe. I am silly, just want to generalize the formula like $$\sum_{j = 1}^n = \dfrac{n(n+1)}{2}$$ the second n+ 1 is upper + lower limit, right? the first n is upper - lower +1 Hence for all lower limit, we just apply it, right?

8. Loser66

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9. Loser66

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10. ParthKohli

can't you apply that in the same way?

11. Loser66

I don't know, I ask and need a confirmation or a logic how to generalize it. Since I don't think we can.

12. ParthKohli

$\sum_{j=m}^{ n}f(j) = \sum_{j=1}^n f(j) - \sum_{j = 1}^{m-1}f(j)$Do you see why the above is true?

13. Loser66

Yes,

14. ParthKohli

Cool. So it can be generalised, right?

15. Loser66

yes

16. ParthKohli

Great.

17. ParthKohli

Just wondering - are you an Indian?

18. Loser66

ok, got it. the first sum = (n(n+1) )/2 the second sum = ((m-1)(m-1+1))/2 some algebraic calculation and we get the answer. Thanks a lot No, I am not. I am Vietnamese

19. imqwerty

sry there was some server update i had to go >.<