1+2+3+4.....+100 =
101*50= 5050
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The mater is that if you do 1+1=2 then just add the sum with its self, so you have 2+2=4 and 4 is the answer
Oh, dude, you're hitting me with the math questions. So, when you add 1 plus 2 plus 3 all the way up to 100, you get... drum roll, please... 5050. Yeah, like, someone actually sat down and figured that out. Crazy, right?
Sum of first n natural numbers is (n) x (n + 1)/2 Here we have the sum = 100 x (101)/2 = 50 x 101 = 5050
Consider a denominator of r; It has proper fractions: 1/r, 2/r, ...., (r-1)/r Their sum is: (1 + 2 + ... + (r-1))/r The numerator of this sum is 1 + 2 + ... + (r-1) Which is an Arithmetic Progression (AP) with r-1 terms, and sum: sum = number_of_term(first + last)/2 = (r-1)(1 + r-1)/2 = (r-1)r/2 So the sum of the proper fractions with a denominator or r is: sum{r} = ((r-1)r/2)/r = ((r-1)r/2r = (r-1)/2 Now consider the sum of the proper fractions with a denominator r+1: sum{r+1} = (((r+1)-1)/2 = ((r-1)+1)/2 = (r-1)/2 + 1/2 = sum{r) + 1/2 So the sums of the proper fractions of the denominators forms an AP with a common difference of 1/2 The first denominator possible is r = 2 with sum (2-1)/2 = ½; The last denominator required is r = 100 with sum (100-1)/2 = 99/2 = 49½; And there are 100 - 2 + 1 = 99 terms to sum So the required sum is: sum = ½ + 1 + 1½ + ... + 49½ = 99(½ + 49½)/2 = 99 × 50/2 = 2475
Oh honey, I'm not a human calculator. But if you want to know the sum of 1 plus 2 repeated until 100, it's 5050. Just add up the numbers from 1 to 100 and voilà, you've got your answer. Math can be a real snooze fest, but hey, at least it's straightforward.