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Another formula for squares

$$\large{\text{Prove that}}$$

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

Note by Trevor Arashiro
1 year, 10 months ago

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Sorry for bad drawing · 1 year, 10 months ago

Nice Megh !

· 1 year, 10 months ago

Impressive! · 1 year, 10 months ago

Thanks dear friend · 1 year, 10 months ago

How do you come up with such beautiful thinking!? I bow before you! Lovely representation! · 1 year, 10 months ago

EXACTLY, This is the pattern that I saw which prompted me to make this note. · 1 year, 10 months ago

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 $$(S_{O} + S_{E}) - (S_{E} - S_{O}) = (2k^{2} + k) - k \Longrightarrow 2S_{O} = 2k^{2} \Longrightarrow S_{O} = k^{2}$$. · 1 year, 10 months ago

Nice one! Did you know this from before of did you just derive this in two minutes? · 1 year, 10 months ago

No, I just saw the pieces, put them together and out came the desired result. :) · 1 year, 10 months ago

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. · 1 year, 10 months ago

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 :-) · 1 year, 10 months ago

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 :) · 1 year, 10 months ago

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 · 1 year, 10 months ago

sorry a small doubt from childhood , seen in many books , what does Q.E.D means? · 1 year, 10 months ago

The "alternative" explanation is "Quite Easily Done". :) Staff · 1 year, 10 months ago

that was a sarcastic remark I guess. · 1 year, 10 months ago

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. Staff · 1 year, 10 months ago