Kofta's blog

1371B — Magical Calender
By Kofta, history, 5 years ago, In English

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 :)

https://www.imageupload.net/image/cf.zXhns

CFbeb06fd0c7aa39c8.png

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5 years ago, # |
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5 years ago, # |
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https://nrich.maths.org/2478

You can find the formula in a manner similar to proving 1+2+3+4+..+n = n*(n+1)/2

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5 years ago, # |
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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

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5 years ago, # |
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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!

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    5 years ago, # ^ |
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    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)

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      5 years ago, # ^ |
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      (b-a+1)*a +(b-a)*(b-a+1)/2 = (b-a+1)*(a + (b-a)/2)
= (b-a+1)*(a+b)/2
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