Problem 4. use the principle of induction to verify: $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}6$$
base case is obviously easy, but I don't know how to prove the inductive case
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Martin Sleziak
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furashu
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See this – David Mitra Jan 31 '14 at 20:37
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1$$\sum_{k=1}^{n+1} k^2 = \left(\sum_{k=1}^n k^2\right) + (n+1)^2.$$ Use the induction hypothesis on the first term. – Daniel Fischer Jan 31 '14 at 20:37
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@DavidMitra thank you. where does the + (m+1)^2 term come from for P(m+1) though? I don't understand that. – furashu Jan 31 '14 at 20:42
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oh duh, that's k^2 = (n+1)^2, we're adding just that last term... – furashu Jan 31 '14 at 20:42
2 Answers
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Hint:
$$\frac{n(n+1)(2n+1)}6+(n+1)^2=(n+1)\left(\frac{n(2n+1)}6+n+1\right)=$$
$$=(n+1)\left(\frac{n(2n+1)+6n+6}6\right)=\frac{n+1}6(n+2)(2n+3)\ldots\ldots$$
DonAntonio
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I'm pretty sure this has been answered before on this site, but anyway here's the proof:
$$\sum_{i=1}^n i^2 + (n+1)^2 = \frac{n(n+1)(2n+1)}{6} + (n+1)^2$$ $$\sum_{i=1}^{n+1} i^2 = \frac{n(n+1)(2n+1) + 6(n+1)^2}{6}$$ $$\sum_{i=1}^{n+1} i^2 = \frac{(n+1)((2n^2+n) + 6(n+1))}{6}$$ $$\sum_{i=1}^{n+1} i^2 = \frac{(n+1)(2n^2+7n+6)}{6}$$ $$\sum_{i=1}^{n+1} i^2 = \frac{(n+1)(2n+3)(n+2)}{6}$$ $$\sum_{i=1}^{n+1} i^2 = \frac{(n+1)(2(n+1)+1)((n+1)+1)}{6}$$
Stefan4024
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I don't understand: how did you get from the third step to the fourth? how can you add those: the first term is in a multiplication right? – furashu Jan 31 '14 at 21:10
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@furashu I missed one parenthesis, maybe that's the reason why you didn't get it. – Stefan4024 Jan 31 '14 at 21:13
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