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shubhamsrg

  • 2 years ago

sum of first natural nos. = n(n+1)/2 my question is that is it just a co-incidence that this sum is also equal to C(n+1 ,2) ? or is there a theoretical explanation for this ?

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  1. shubhamsrg
    • 2 years ago
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    i mean can we arrive at sum of n digits formula using permutation/combination/binomial ?

  2. KingGeorge
    • 2 years ago
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    Well, recursively, you can see that \[\binom{n+1}{2}=\binom{n}{2}+\binom{n}{1}=\sum_{i=0}^{n-1}i+n\]So the result follows easily from induction.

  3. ParthKohli
    • 2 years ago
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    \[C(n + 1, 2) = \dfrac{(n + 1)!}{(n - 1)!2!} = \dfrac{(n + 1)n\cancel{(n -1)!}}{\cancel{(n -1)!}2!}\]

  4. KingGeorge
    • 2 years ago
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    Also, suppose you have \(n+1\) objects to choose 2 things from, but order doesn't matter. Then you have \(n+1\) choices for the first thing, and \(n\) choices for the second. but since order doesn't matter, you have to divide by \(2!=2\). Resulting in \[\frac{(n+1)\cdot n}{2}\]Basically what parth just did above me.

  5. shubhamsrg
    • 2 years ago
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    we are selecting 2 objects from n+1 things, okay, what do you mean by having n+1 choices for first thing ?

  6. ParthKohli
    • 2 years ago
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    And then there's a proof without words:

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  7. KingGeorge
    • 2 years ago
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    We have a total of \(n+1\) objects, and we can choose any of them for our first choice. Thus, we have a total of \(n+1\) choices since every object can be chosen.

  8. shubhamsrg
    • 2 years ago
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    hmm so we have n+1 choices for first one n for second so total combinations = n(n+1)/2! okay, fair enough how you relate that to sum of n natural nos. ?

  9. KingGeorge
    • 2 years ago
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    That explanation isn't very well connected to the sum of n natural numbers. However, the first explanation that I gave is probably a better explanation of what you're looking for. The base case is \(\displaystyle\binom{2}{2}\), and after that, the result follows from the definition of \(\displaystyle\binom{n}{k}\).

  10. KingGeorge
    • 2 years ago
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    I don't have any solid explanation of any inherent reason that \(\binom{2}{2}\) is the same as the sum of the natural numbers starting and ending at 1, other than coincidence.

  11. shubhamsrg
    • 2 years ago
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    C(n,2) = 1+2+3...(n-1) -->am stuck here..

  12. shubhamsrg
    • 2 years ago
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    thats actually my question only if you replace n by n-1 .. :P

  13. shubhamsrg
    • 2 years ago
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    i mean n by n+1

  14. KingGeorge
    • 2 years ago
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    The first proof I gave shows that. Take 5-choose-2 for example.\[\binom{5}{2}=\binom{4}{2}+\binom41=\binom32+\binom31+4=\binom22+\binom21+3+4=1+2+3+4\]

  15. shubhamsrg
    • 2 years ago
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    ohh..got it!! :O

  16. shubhamsrg
    • 2 years ago
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    should have understood in the first attempt only..hmm my bad! thanks again..! :)

  17. ParthKohli
    • 2 years ago
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    @KingGeorge: Quick question... can you use the Binomial Theorem here? ;)

  18. KingGeorge
    • 2 years ago
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    I'm not sure. The issue with the binomial theorem, is that you get every \(\binom{n}{k}\) for \(0\le k\le n\).

  19. ParthKohli
    • 2 years ago
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    I see that there's a really cool proof to \(\binom{n}{0} +\binom{n}{1}+ \binom{n}{2} \cdots\binom{n}{n} = 2^n\)

  20. UnkleRhaukus
    • 2 years ago
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    \[ \stackrel{n+1}{\overbrace{\left.\begin{array}{ccccc}0&0&0&0&\cdots &0\\&1&1&1&\cdots&1\\&&2&2&\cdots&2\\&&&3&\cdots&3\\&&&&\ddots&\vdots\\&&&&&n\end{array}\right\}n}{}}\] \[A_\triangle=\frac{hb}2=\frac{n(n+1)}{2}\]

  21. ParthKohli
    • 2 years ago
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    Wow!!

  22. shubhamsrg
    • 2 years ago
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    i didnt understand @UnkleRhaukus area is okay,how do you relate sum of digits ?

  23. ParthKohli
    • 2 years ago
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    @shubhamsrg: \(\underbrace{1 + 2 + 3 + 4 \cdots n}_{h} + \underbrace{0 + 0 + 0\cdots}_{b}\)

  24. KingGeorge
    • 2 years ago
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    If we take \[(0+1)^2=\binom200^21^0+\binom210^11^1+\binom220^01^2=0+0+\binom22\]\[(0+1)^2=1^2=1\]So \[\binom22=1\]But I don't really see how this helps with that issue.

  25. KingGeorge
    • 2 years ago
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    Also, @UnkleRhaukus, there are n+1 rows in that diagram, not n.

  26. shubhamsrg
    • 2 years ago
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    so you;re saying area = (1+2+3..n) + (0+0...(n+1 times) )

  27. UnkleRhaukus
    • 2 years ago
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    oh bother

  28. ParthKohli
    • 2 years ago
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    But the geometric proofs are really pretty. =)

  29. shubhamsrg
    • 2 years ago
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    i did not, follow the geometric proof .. ?

  30. shubhamsrg
    • 2 years ago
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    by not follow i mean didnt get*

  31. shubhamsrg
    • 2 years ago
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    @UnkleRhaukus ? @ParthKohli ?

  32. ParthKohli
    • 2 years ago
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    When you calculate the area of ANY figure, you count the number of units it covers. So just add the number of units (numbers) inside it: \((n + 1) + n + (n - 1) + (n - 2)\cdots 1 \) which is just the sum of first \(n + 1\) natural numbers. But the area ALSO equals base times height, so...

  33. ParthKohli
    • 2 years ago
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    half the base times height*

  34. shubhamsrg
    • 2 years ago
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    got it! B|

  35. shubhamsrg
    • 2 years ago
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    that was really cool..thanks @UnkleRhaukus @ParthKohli and surely @KingGeorge

  36. KingGeorge
    • 2 years ago
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    You're very welcome.

  37. ParthKohli
    • 2 years ago
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    :)

  38. shubhamsrg
    • 2 years ago
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    ohh wait

  39. ParthKohli
    • 2 years ago
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    Yes, correction: it's the sum of the first \(n\) natural numbers.

  40. shubhamsrg
    • 2 years ago
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    (n+1) + n + n-1 ... 1 = (n+2)(n+1)/2 also, in the figure, height = n+1

  41. ParthKohli
    • 2 years ago
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    Yes, so the height is just the first \(h\) natural numbers.

  42. ParthKohli
    • 2 years ago
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    But if we assume that \(n +1 = h\), then we have \(h(h+1)/2\)

  43. shubhamsrg
    • 2 years ago
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    height = n+1 base =n +1 so area accordingly = (n+1)^2 /2

  44. shubhamsrg
    • 2 years ago
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    0,1,2...n -->n+1 units 0 also written n+1 times.. hmm ?

  45. ParthKohli
    • 2 years ago
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    Yeah...

  46. ParthKohli
    • 2 years ago
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    But @KingGeorge pointed out a mistake earlier. Not sure how I should comprehend the diagram ._.

  47. UnkleRhaukus
    • 2 years ago
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    \[\color{red}*\]\[\stackrel{n+1}{\left.\overbrace{\begin{array}{ccccc}0&0&0&0&\cdots &0\\&1&1&1&\cdots&1\\&&2&2&\cdots&2\\&&&3&\cdots&3\\&&&&\ddots&\vdots\\&&&&&n-1\end{array}{}}\right\}}\small n\]

  48. UnkleRhaukus
    • 2 years ago
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    the sum of the first \(n\)-natural numbers starting at (n=1,0)

  49. shubhamsrg
    • 2 years ago
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    no..following the pattern, n-1 should have been written 2 times..

  50. shubhamsrg
    • 2 years ago
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    0 is written n+1 times 1 -> n times =>sum is n+1 2->n-1 times =>sum is n+1 . . .n-1 -> 2 times =>sum should be n+1

  51. shubhamsrg
    • 2 years ago
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    so whats the flaw in geometrical proof ? please get back to this..

  52. shubhamsrg
    • 2 years ago
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    as the geometrical proof seems correct, but shouldnt have been correct! :|

  53. KingGeorge
    • 2 years ago
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    I would like to take a moment and go back to the "proof without words" @ParthKohli posted. I just realized exactly what that was showing, and it draws a very nice bijection between the sum of the natural numbers and C(n,2). The bottom line has n dots, and above, you have 1+2+...+n-1 dots. Now, choose two dots on the bottom line, and draw two diagonal lines up so that they meet somewhere above. These lines will intersect in another dot. Similarly, if you choose some dot not in the bottom line, and draw two diagonal lines downward. Each of these lines will hit some dot in the bottom row and you get two dots of the n dots. This demonstrates a function with an inverse, so it's a bijection, and the sum of the natural numbers up to n-1 is the same as C(n,2).

  54. KingGeorge
    • 2 years ago
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    The advantage of this method, is that it takes care of the C(2,2) case that I had issues with because when you have 2 dots in the bottom row, there's still a row above with 1 dot, so C(2,2)=1.

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