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

Question

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

parthkohli · Answer

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

parthkohli · Answer

sorry for lack of notation

imqwerty · Answer

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

imqwerty · Answer

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

loser66 · Answer

Got it. thank you.

imqwerty · Answer

np :)

loser66 · Answer

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?

loser66 · Answer

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