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) = \(k^{2}\) + 2(k+1) -1 = \(k^{2}\) +2k +1 = \((k+1)^{2}\), as required. Q.E.D.

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.

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

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

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## Comments

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TopNewestSorry for bad drawing

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Nice Megh !

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Impressive!

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Thanks dear friend

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

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Sorry for late response , thought about this long back

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

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

Then \(S_{E} - S_{O} = k\) and \(S_{O} + S_{E} = \displaystyle\sum_{n=1}^{2k} n = \dfrac{2k(2k + 1)}{2} = 2k^{2} + k\).

Therefore \((S_{O} + S_{E}) - (S_{E} - S_{O}) = (2k^{2} + k) - k \Longrightarrow 2S_{O} = 2k^{2} \Longrightarrow S_{O} = k^{2}\).

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

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

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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) = \(k^{2}\) + 2(k+1) -1 = \(k^{2}\) +2k +1 = \((k+1)^{2}\), as required. Q.E.D.

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starting with n=1, (2

1 - 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.DLog in to reply

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

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It's

quaderademonstrandumwhich means"which was to be demonstrated"Log in to reply

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

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Q.E.D. Modern humorous usage

It's a running joke in the Math community. See for example, WikipediaNote: I generally avoid making sarcastic remarks over the internet, unless the context is very easily deduced. his is because sacarsm doesn't transfer well.

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

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

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