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

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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

Mathematics
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\[= \sum_{1}^{n}j - \sum_{1}^{m-1}j\]
sorry for lack of notation
terms from m to n = n-m+1 so summation wuld be- \[\frac{ (n-m+1)(m+n) }{ 2 }\]

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number of terms =n-m+1 1st term =m and common diff=1 so just put this in the formula of summation of ap :)
Got it. thank you.
np :)
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?
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can't you apply that in the same way?
I don't know, I ask and need a confirmation or a logic how to generalize it. Since I don't think we can.
\[\sum_{j=m}^{ n}f(j) = \sum_{j=1}^n f(j) - \sum_{j = 1}^{m-1}f(j)\]Do you see why the above is true?
Yes,
Cool. So it can be generalised, right?
yes
Great.
Just wondering - are you an Indian?
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
sry there was some server update i had to go >.<

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