### MostafaKeller's blog

By MostafaKeller, history, 6 weeks ago, ,

I can't understand the line that covered by red. also can someone tell me the formula's name or it's proof.

Thanks for any future comments :)

• 0

 » 6 weeks ago, # | ← Rev. 2 →   0 I can't delete this comment...
 » 6 weeks ago, # |   0 https://nrich.maths.org/2478You can find the formula in a manner similar to proving 1+2+3+4+..+n = n*(n+1)/2
•  » » 6 weeks ago, # ^ |   0 Thanks
 » 6 weeks ago, # |   0 This is an arithmetic series. usually you can evaluate sequences x, x+d, x+2d, ... x+(n-1)*d by using the fact that x + x+(n-1)*d (simply pairing up the first and nth terms) equals x+d + x+(n-2)*d, + etc so basically you can just find the average of each element, which is (x+x+(n-1)*d) /2 * number of terms
•  » » 6 weeks ago, # ^ |   0 Thanks
 » 6 weeks ago, # |   0 a + (a+1) + (a+2) ... + (b) --> b-a+1 terms This evaluates to = (b-a+1)*a + (1 + 2 + 3 .. b-a) = (b-a+1)*a + (b-a)*(b-a+1)/2 = (b-a+1)*(a+b)/2 Hope it helps!
•  » » 6 weeks ago, # ^ |   0 In second line when we take (b-a+1) as common factor isn't it look like that (b-a+1)*(a+b-a)/2 ? how does it become (a+b)
•  » » » 6 weeks ago, # ^ |   0 (b-a+1)*a +(b-a)*(b-a+1)/2 = (b-a+1)*(a + (b-a)/2) = (b-a+1)*(a+b)/2 
•  » » » » 6 weeks ago, # ^ |   0 Ah it's clear now Thanks