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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 ?

Mathematics
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i mean can we arrive at sum of n digits formula using permutation/combination/binomial ?
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.
\[C(n + 1, 2) = \dfrac{(n + 1)!}{(n - 1)!2!} = \dfrac{(n + 1)n\cancel{(n -1)!}}{\cancel{(n -1)!}2!}\]

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Other answers:

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.
we are selecting 2 objects from n+1 things, okay, what do you mean by having n+1 choices for first thing ?
And then there's a proof without words:
1 Attachment
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.
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. ?
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 \(\displaystyle\binom{n}{k}\).
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.
C(n,2) = 1+2+3...(n-1) -->am stuck here..
thats actually my question only if you replace n by n-1 .. :P
i mean n by n+1
The first proof I gave shows that. Take 5-choose-2 for example.\[\binom{5}{2}=\binom{4}{2}+\binom41=\binom32+\binom31+4=\binom22+\binom21+3+4=1+2+3+4\]
ohh..got it!! :O
should have understood in the first attempt only..hmm my bad! thanks again..! :)
@KingGeorge: Quick question... can you use the Binomial Theorem here? ;)
I'm not sure. The issue with the binomial theorem, is that you get every \(\binom{n}{k}\) for \(0\le k\le n\).
I see that there's a really cool proof to \(\binom{n}{0} +\binom{n}{1}+ \binom{n}{2} \cdots\binom{n}{n} = 2^n\)
\[ \stackrel{n+1}{\overbrace{\left.\begin{array}{ccccc}0&0&0&0&\cdots &0\\&1&1&1&\cdots&1\\&&2&2&\cdots&2\\&&&3&\cdots&3\\&&&&\ddots&\vdots\\&&&&&n\end{array}\right\}n}{}}\] \[A_\triangle=\frac{hb}2=\frac{n(n+1)}{2}\]
Wow!!
i didnt understand @UnkleRhaukus area is okay,how do you relate sum of digits ?
@shubhamsrg: \(\underbrace{1 + 2 + 3 + 4 \cdots n}_{h} + \underbrace{0 + 0 + 0\cdots}_{b}\)
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.
Also, @UnkleRhaukus, there are n+1 rows in that diagram, not n.
so you;re saying area = (1+2+3..n) + (0+0...(n+1 times) )
oh bother
But the geometric proofs are really pretty. =)
i did not, follow the geometric proof .. ?
by not follow i mean didnt get*
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...
half the base times height*
got it! B|
that was really cool..thanks @UnkleRhaukus @ParthKohli and surely @KingGeorge
You're very welcome.
:)
ohh wait
Yes, correction: it's the sum of the first \(n\) natural numbers.
(n+1) + n + n-1 ... 1 = (n+2)(n+1)/2 also, in the figure, height = n+1
Yes, so the height is just the first \(h\) natural numbers.
But if we assume that \(n +1 = h\), then we have \(h(h+1)/2\)
height = n+1 base =n +1 so area accordingly = (n+1)^2 /2
0,1,2...n -->n+1 units 0 also written n+1 times.. hmm ?
Yeah...
But @KingGeorge pointed out a mistake earlier. Not sure how I should comprehend the diagram ._.
\[\color{red}*\]\[\stackrel{n+1}{\left.\overbrace{\begin{array}{ccccc}0&0&0&0&\cdots &0\\&1&1&1&\cdots&1\\&&2&2&\cdots&2\\&&&3&\cdots&3\\&&&&\ddots&\vdots\\&&&&&n-1\end{array}{}}\right\}}\small n\]
the sum of the first \(n\)-natural numbers starting at (n=1,0)
no..following the pattern, n-1 should have been written 2 times..
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
so whats the flaw in geometrical proof ? please get back to this..
as the geometrical proof seems correct, but shouldnt have been correct! :|
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).
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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