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

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shubhamsrgBest ResponseYou've already chosen the best response.2
i mean can we arrive at sum of n digits formula using permutation/combination/binomial ?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.6
Well, recursively, you can see that \[\binom{n+1}{2}=\binom{n}{2}+\binom{n}{1}=\sum_{i=0}^{n1}i+n\]So the result follows easily from induction.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
\[C(n + 1, 2) = \dfrac{(n + 1)!}{(n  1)!2!} = \dfrac{(n + 1)n\cancel{(n 1)!}}{\cancel{(n 1)!}2!}\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.6
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.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
we are selecting 2 objects from n+1 things, okay, what do you mean by having n+1 choices for first thing ?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
And then there's a proof without words:
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.6
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.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
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. ?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.6
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 \(\displaystyle\binom{n}{k}\).
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.6
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.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
C(n,2) = 1+2+3...(n1) >am stuck here..
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
thats actually my question only if you replace n by n1 .. :P
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.6
The first proof I gave shows that. Take 5choose2 for example.\[\binom{5}{2}=\binom{4}{2}+\binom41=\binom32+\binom31+4=\binom22+\binom21+3+4=1+2+3+4\]
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
should have understood in the first attempt only..hmm my bad! thanks again..! :)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
@KingGeorge: Quick question... can you use the Binomial Theorem here? ;)
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.6
I'm not sure. The issue with the binomial theorem, is that you get every \(\binom{n}{k}\) for \(0\le k\le n\).
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
I see that there's a really cool proof to \(\binom{n}{0} +\binom{n}{1}+ \binom{n}{2} \cdots\binom{n}{n} = 2^n\)
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.2
\[ \stackrel{n+1}{\overbrace{\left.\begin{array}{ccccc}0&0&0&0&\cdots &0\\&1&1&1&\cdots&1\\&&2&2&\cdots&2\\&&&3&\cdots&3\\&&&&\ddots&\vdots\\&&&&&n\end{array}\right\}n}{}}\] \[A_\triangle=\frac{hb}2=\frac{n(n+1)}{2}\]
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
i didnt understand @UnkleRhaukus area is okay,how do you relate sum of digits ?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
@shubhamsrg: \(\underbrace{1 + 2 + 3 + 4 \cdots n}_{h} + \underbrace{0 + 0 + 0\cdots}_{b}\)
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.6
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.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.6
Also, @UnkleRhaukus, there are n+1 rows in that diagram, not n.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
so you;re saying area = (1+2+3..n) + (0+0...(n+1 times) )
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
But the geometric proofs are really pretty. =)
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
i did not, follow the geometric proof .. ?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
by not follow i mean didnt get*
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
@UnkleRhaukus ? @ParthKohli ?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
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...
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
half the base times height*
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
that was really cool..thanks @UnkleRhaukus @ParthKohli and surely @KingGeorge
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.6
You're very welcome.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Yes, correction: it's the sum of the first \(n\) natural numbers.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
(n+1) + n + n1 ... 1 = (n+2)(n+1)/2 also, in the figure, height = n+1
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Yes, so the height is just the first \(h\) natural numbers.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
But if we assume that \(n +1 = h\), then we have \(h(h+1)/2\)
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
height = n+1 base =n +1 so area accordingly = (n+1)^2 /2
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
0,1,2...n >n+1 units 0 also written n+1 times.. hmm ?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
But @KingGeorge pointed out a mistake earlier. Not sure how I should comprehend the diagram ._.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.2
\[\color{red}*\]\[\stackrel{n+1}{\left.\overbrace{\begin{array}{ccccc}0&0&0&0&\cdots &0\\&1&1&1&\cdots&1\\&&2&2&\cdots&2\\&&&3&\cdots&3\\&&&&\ddots&\vdots\\&&&&&n1\end{array}{}}\right\}}\small n\]
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.2
the sum of the first \(n\)natural numbers starting at (n=1,0)
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
no..following the pattern, n1 should have been written 2 times..
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
0 is written n+1 times 1 > n times =>sum is n+1 2>n1 times =>sum is n+1 . . .n1 > 2 times =>sum should be n+1
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
so whats the flaw in geometrical proof ? please get back to this..
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
as the geometrical proof seems correct, but shouldnt have been correct! :
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.6
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+...+n1 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 n1 is the same as C(n,2).
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.6
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.
 one year ago
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