Prove that: $$ 9+16+\dots+(n+2)^2 = \frac{n^3}{3}+\frac{5n^2}{2}+\frac{37n}{6} $$with induction.
I managed to prove the base step but got stuck with the induction step.
Could anyone please help me with the induction step?
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2Just copy the induction proof for $1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}$ and rewrite. – Dietrich Burde Feb 21 '20 at 12:05
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Will take a look at it, thank you for the link. – Leon Feb 21 '20 at 12:11
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Cf. this question – J. W. Tanner Feb 21 '20 at 13:17
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But my answer is still okay, right! – Nεo Pλατo Feb 21 '20 at 14:30
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@NeoPlato Yes. Thank you! – Leon Feb 21 '20 at 16:12
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Any time man. I'm also practicing so thanks too. – Nεo Pλατo Feb 21 '20 at 16:57
1 Answers
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For the inductive step all you do is show that if the formula is true for $n$ it is true for $n+1$. In this case, you show that taking the value for $n$ and adding $n+1$ is the same as using $n+1$ as the input:
$\frac{n^3}{3}+\frac{5n^2}{2}+\frac{37n}{6}+((n+1)+2)^2=\frac{(n+1)^3}{3}+\frac{5(n+1)^2}{2}+\frac{37(n+1)}{6}$
$\frac{n^3}{3}+\frac{5n^2}{2}+\frac{37n}{6}+((n+1)+2)^2=\frac{n^3}{3}+\frac{5n^2}{2}+\frac{37n}{6}+n^2+6n+9$
$=\frac{n^3}{3}+n^2+n+\frac{1}{3}+\frac{5n^2}{2}+5n+\frac{5}{2}+\frac{37n}{6}+\frac{37}{6} \implies \text{Distribution of extra terms.}$
$=\frac{n^3+3n^2+3n+1}{3}+\frac{5n^2+10n+5}{2}+\frac{37n+37}{6}$
$=\frac{(n+1)^3}{3}+\frac{5(n+1)^2}{2}+\frac{37(n+1)}{6}$
So if it works for 1, it works for 2, 3 and so on, "which is the very thing it was required to show". $\implies$ Elements, Euclid
Nεo Pλατo
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