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i mean can we arrive at sum of n digits formula using permutation/combination/binomial ?

And then there's a proof without words:

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

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? ;)

Wow!!

i didnt understand @UnkleRhaukus
area is okay,how do you relate sum of digits ?

@shubhamsrg: \(\underbrace{1 + 2 + 3 + 4 \cdots n}_{h} + \underbrace{0 + 0 + 0\cdots}_{b}\)

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*

half the base times height*

got it! B|

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 ._.

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..

so whats the flaw in geometrical proof ?
please get back to this..

as the geometrical proof seems correct, but shouldnt have been correct! :|