Another formula for squares

Prove that\large{\text{Prove that}}

n=1k(2n1)=k2\large{\displaystyle \sum_{n=1}^k (2n-1)=k^2}

Note by Trevor Arashiro
4 years, 11 months ago

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Sorry for bad drawing

U Z - 4 years, 11 months ago

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Nice Megh !

Deepanshu Gupta - 4 years, 10 months ago

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Impressive!

Vraj Mehta - 4 years, 11 months ago

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Thanks dear friend

U Z - 4 years, 11 months ago

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How do you come up with such beautiful thinking!? I bow before you! Lovely representation!

Kunal Jadhav - 4 years, 10 months ago

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Sorry for late response , thought about this long back

U Z - 4 years, 10 months ago

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EXACTLY, This is the pattern that I saw which prompted me to make this note.

Trevor Arashiro - 4 years, 11 months ago

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Let SO=n=1k(2n1)S_{O} = \displaystyle\sum_{n=1}^{k} (2n - 1) and SE=n=1k(2n)S_{E} = \displaystyle\sum_{n=1}^{k} (2n).

Then SESO=kS_{E} - S_{O} = k and SO+SE=n=12kn=2k(2k+1)2=2k2+kS_{O} + S_{E} = \displaystyle\sum_{n=1}^{2k} n = \dfrac{2k(2k + 1)}{2} = 2k^{2} + k.

Therefore (SO+SE)(SESO)=(2k2+k)k2SO=2k2SO=k2(S_{O} + S_{E}) - (S_{E} - S_{O}) = (2k^{2} + k) - k \Longrightarrow 2S_{O} = 2k^{2} \Longrightarrow S_{O} = k^{2}.

Brian Charlesworth - 4 years, 11 months ago

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Nice one! Did you know this from before of did you just derive this in two minutes?

Trevor Arashiro - 4 years, 11 months ago

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No, I just saw the pieces, put them together and out came the desired result. :)

Brian Charlesworth - 4 years, 11 months ago

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I will use proof by induction to show that if the formula is true for p(k) then it is true for p(k+1) and hence it is true for all natural numbers. Let s(k) denote the sum for values up to k. Base step: s(1) = 1, s(2) = 1+3 = 4 ... so the formula is true so far, as no counter-examples have been found. Inductive step: s(k+1) = k2k^{2} + 2(k+1) -1 = k2k^{2} +2k +1 = (k+1)2(k+1)^{2}, as required. Q.E.D.

Curtis Clement - 4 years, 11 months ago

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starting with n=1, (21 - 1) + (22-1) + (23-1) +. . . . . . . . . (2k -1) 21 -1+ 22 - 1 +23 -1 +.............. 2k - 1 22 + 22 + 23 ....2k - k1 2(1 + 2+ 3+ ......k) - k using AP sum formula;{1 + 2+ 3+...k= k/2(21 +k-1) = k/2(1+k) keeping values, =2*k/2(1+k) - k = k^2 Q.E.D

Raven Herd - 4 years, 11 months ago

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sorry a small doubt from childhood , seen in many books , what does Q.E.D means?

U Z - 4 years, 11 months ago

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It's quad era demonstrandum which means "which was to be demonstrated"

Kushal Patankar - 4 years, 10 months ago

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The "alternative" explanation is "Quite Easily Done". :)

Calvin Lin Staff - 4 years, 10 months ago

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@Calvin Lin that was a sarcastic remark I guess.

Raven Herd - 4 years, 10 months ago

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@Raven Herd It's a running joke in the Math community. See for example, Wikipedia Q.E.D. Modern humorous usage

Q.E.D. is sometimes jokingly claimed to abbreviate "quite easily done", or "Quit. Enough done."

Note: I generally avoid making sarcastic remarks over the internet, unless the context is very easily deduced. his is because sacarsm doesn't transfer well.

Calvin Lin Staff - 4 years, 10 months ago

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You could also do it from r2(r1)2=2r1r^2-(r-1)^2=2r-1 and them sum both sides from r=1r=1 to r=kr=k. The LHS just telescopes and gives you k2k^2 and the RHS becomes the LHS of the given q :-)

Daniel Remo - 4 years, 10 months ago

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I agree that it would provide a nice alternative as i=1n\displaystyle \sum_{i=1}^n r2r^{2} - (r1)2(r-1)^{2} = n(n+1)(2n+1)6\frac{n(n+1)(2n+1)}{6} - n(n1)(2n1)6\frac{n(n-1)(2n-1)}{6} = n6\frac{n}{6} [(2n2n^{2} +3n{n} +1) - ((2n2n^{2} -3n{n} +1)] = n2n^{2}, which is less wordy :)

Curtis Clement - 4 years, 10 months ago

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