Problem: Prove that $1^2+2^2+cdots+n^2=fracn(n+1)(2n+1)6$ for $n in igbiglands.combbN$.

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My work: So I think I have to do a proof by induction và I just wanted some help editing my proof.

My attempt:

Let $P(n)=1^2+2^2+cdots+n^2=fracn(n+1)(2n+1)6$ for $n in igbiglands.combbN$. Then $$P(1)=1^2=frac1(1+1)(2+1)6$$$$1=frac66.$$So $P(1)$ is true.

Next suppose that $P(k)=1^2+2^2+cdots+k^2=frack(k+1)(2k+1)6$ for $k in igbiglands.combbN$. Then adding $(k+1)^2$ lớn both sides of $P(k)$ we obtain the following:$$1^2+2^2+cdots+k^2+(k+1)^2=frack(k+1)(2k+1)6+(k+1)^2$$$$=frac2k^3+3k^2+k+6(k^2+2k+1)6$$$$=frac2k^3+9k^2+13k+66$$$$=frac(k^2+3k+2)(2k+3)6$$$$=frac(k+1)(k+2)(2k+3)6$$$$=frac(k+1)((k+1)+1)(2(k+1)+1)6$$$$=P(k+1).$$Thus $P(k)$ is true for $k in igbiglands.combbN$.Hence by bigbiglands.comematical induction, $1^2+2^2+cdots+n^2=fracn(n+1)(2n+1)6$ is true for $n in igbiglands.combbN$.

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edited Feb 26 "15 at 6:22

Daniel W. Farlow
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asked Feb 26 "15 at 5:28
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$egingroup$ Oh yes, would you like to know? $endgroup$
Feb 26 "15 at 5:41
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3 Answers 3

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Consider any natural number $r$. You have $$r^3-(r-1)^3=3r^2-3r+1.$$

Now telescope it: $$1^3-0^3=3-3+1$$$$2^3-1^3=3cdot2^2-3cdot2+1$$$$vdots$$$$n^3-(n-1)^3=3n^2-3n+1$$ Now add, and see them cancel out. You are left with $$n^3=3(1^2+2^2+cdots+ n^2)-3(1+2+3+cdots+n)+n$$ You must know $$1+2+3+cdots+n=fracn(n+1)2.$$ Plug it in, and you get the answer. Also, please see that this method works even for $sum r^4,r^5,cdots$. I have tried it out. All you need is the sum of its previous powers.

nói qua
edited Mar 6 "15 at 9:49
answered Feb 26 "15 at 5:48
$egingroup$ Oh, I too thought it looked bad. Thanks. $endgroup$
Feb 26 "15 at 6:06
$egingroup$ Hope it is better now. $endgroup$
Feb 26 "15 at 6:09
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I am going to provide what I think is a nice way of writing up a proof, both in terms of accuracy & in terms of communication. You be the judge(s).

Claim: For $ngeq 1$, let $S(n)$ be the statement$$S(n) : 1^2+2^2+3^2+cdots+n^2=fracn(n+1)(2n+1)6.$$

Base step $(n=1)$: The statement $S(1)$ says $1^2=1(2)(3)/6$ which is true.

Inductive step $(S(k) o S(k+1))$: Fix some $kgeq 1$ & suppose that$$S(k) : 1^2+2^2+3^2+cdots+k^2=frack(k+1)(2k+1)6$$holds. Lớn be shown is that$$S(k+1) : 1^2+2^2+3^2+cdots+k^2+(k+1)^2=frac(k+1)(k+2)(2(k+1)+1)6$$follows. Starting with the left-hand side of $S(k+1)$,eginalign extLHS &= 1^2+2^2+3^2+cdots+k^2+(k+1)^2 agdefinition\<1em> &= frack(k+1)(2k+1)6+(k+1)^2 agby $S(k)$\<1em> &= (k+1)left\<1em> &= (k+1)frack(2k+1)+6(k+1)6\<1em> &= (k+1)frac2k^2+k+6k+66\<1em> &= (k+1)frac2k^2+7k+66\<1em> &= (k+1)frac(k+2)(2k+3)6\<1em> &= frac(k+1)(k+2)(2(k+1)+1)6\<1em> &= extRHS,endalignthe right-hand side of $S(k+1)$ follows. This completes the inductive step.

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