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  • Proof that $1+2+3+4+\cdots+n = \frac {n\times (n+1)}2$
    Why is $1+2+3+4+\ldots+n = \dfrac {n\times (n+1)}2$ $\space$ ? Not appropriate for an answer, but you've asked either a very easy or a very difficult question If by "why" you mean, "Can I see a proof of this fact?" the question is fairly easy to answer If by "why" you mean, "Why should this be true?" you've asked a very deep kind of question that mathematicians make entire careers out of
  • combinatorics - Exactly how does the equation n (n-1) 2 determine the . . .
    Exactly how does the equation n (n-1) 2 determine the number of pairs of a given number of items (n)? Ask Question Asked 9 years, 2 months ago Modified 4 years, 7 months ago
  • Dont understand why N * (N -1 ) 2 : r mathematics - Reddit
    Repeat the process until your list is empty - you now have N 2 pairs of numbers that each add to N+1 Hence, the sum of all integers from 1 to an even N is (N+1)*N 2 It’s a couple steps more to show that this also works for odd N, and that you get the formula you asked about if you replace N with N-1
  • how n * (n + 1) 2 does summation? : r askmath - Reddit
    The rectangle will have dimensions (n+1) by n, and it uses 2 “staircases” so dividing by 2 gives that each staircase, which has a number of blocks equal to the sum of the first n integers, has n (n+1) 2 blocks
  • formula - What is the proof of of (N–1) + (N–2) + (N–3) + . . . + 1= N . . .
    This is an arithmetic series, and the equation for the total number of times is (n - 1)*n 2 Example: if the size of the list is N = 5, then you do 4 + 3 + 2 + 1 = 10 swaps -- and notice that 10 is the same as 4 * 5 2
  • math - Is 2^ (2n) = O (2^n) - Stack Overflow
    O (2^ (n+1)) is the same as O (2 * 2^n), and you can always pull out constant factors, so it is the same as O (2^n) However, constant factors are the only thing you can pull out 2^ (2n) can be expressed as (2^n) (2^n), and 2^n isn't a constant
  • Why sum of integer from 1 to n is (n* (n+1)) 2 ?? : r learnmath - Reddit
    Then you are proving that if you take an arbitrary sum from 1 to n, that if that holds, it also holds for the next element In this case it means that Sum (1 to n+1) = Sum (1 to n) + (n+1)
  • Rule of exponents -- why does $2^n + 2^n = 2^ {n+1}$
    Rule of exponents -- why does $2^n + 2^n = 2^ {n+1}$ Ask Question Asked 12 years, 2 months ago Modified 11 years, 7 months ago Viewed 2k times
  • Big O, what is the complexity of summing a series of n numbers?
    It would help to know what "this" is You're right that adding up n things (doing something n times, each of cost O (1)) is O (n) But if instead of adding 1+2+3+ etc you had to do something once, and then do something twice, and then three times, etc , then after 1+2+3 +n were done you'd have done n* (n+1) 2 things, which is O (n^2)
  • Is there a name for the formula (n* (n+1)) 2 : r learnmath - Reddit
    Basically, (n* (n+1)) 2 = 1+2+3+…n Like a factorial but with addition instead of multiplication I’ve been googling trying to find a name for it just so i can describe it better but everywhere I look doesn’t seem to give a straight answer I’ve seen some people calling it a triangular number, due to the shape it takes when you write it out Some other people are calling it just a





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