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\$.

proof-verification proof-writing induction
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edited Feb 26 "15 at 6:22

Daniel W. Farlow
asked Feb 26 "15 at 5:28

bigbiglands.com Majorbigbiglands.com Major
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\$egingroup\$ Oh yes, would you like to know? \$endgroup\$
–user167045
Feb 26 "15 at 5:41

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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.

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edited Mar 6 "15 at 9:49
answered Feb 26 "15 at 5:48
user167045user167045
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\$egingroup\$ Oh, I too thought it looked bad. Thanks. \$endgroup\$
–user167045
Feb 26 "15 at 6:06
\$egingroup\$ Hope it is better now. \$endgroup\$
–user167045
Feb 26 "15 at 6:09