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I need to prove $$\sum_{k=1}^n(2k)^2 = \frac{2n(n+1)(2n+1)}{3}$$ using mathematical induction. I tried solving this but I got stuck when proving that this is true for n=k+1.

I would be very thankful if someone could help me.

Carolina
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  • Do you know the formula for $\displaystyle\sum_{k=1}^n k^2$? – Bernard Sep 10 '17 at 15:16
  • We would be better off without assuming we know $$\sum_{k=1}^{n} k^2=\frac{n(n+1)(2n+1)}{6}$$ since this is an exercise regarding induction. – Deniz Tuna Yalçın Sep 10 '17 at 15:43
  • I didn't know the formula for $\sum_{k=1}^n k^2$ before. But it helps to solve this easily. Thanks. – Carolina Sep 10 '17 at 17:21
  • You're right it makes things easier in terms of checking whether The formula is true. İt helps you know if you're making a mistake and that is important because sometimes The formulas are not right. By knowing they are true inspired from a commonly-known formula is a luxury though we shouldn't get used to this. And if we are to add them in our solutions we have to prove them. – Deniz Tuna Yalçın Sep 10 '17 at 17:44

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Let's call this expression $\phi(n)$ if $\phi(n)$ is true than $\phi(n+1)$ should also be true

$\phi(n)$ generates your formula above. $$\phi(n+1)=\sum_{k=1}^{n+1}(2k)^2=\sum_{k=1}^{n}(2k)^2+2(n+1)^2$$ $$=\frac{2n\cdot(n+1)(2n+1)}{3}+2(n+1)^2$$ This is the expression we have to check and we have to check that it equals what we have when we plug in $n=n+1$ for the above formula so; $$\frac{2n\cdot(n+1)(2n+1)}{3}+2(n+1)^2=\frac{2(n+1)((n+1)+1)(2(n+1)+1)}{3}$$ Let's workout the operations;$$\frac{2n(n+1)(2n+1)+6(n+1)^2}{3}$$ $$\frac{(n+1)(4n^2+2n+6n+6)}{3}$$ $$\frac{2(n+1)(2n^2+4n+3)}{3}$$ $$\frac{2(n+1)(n+2)(2n+3)}{3}$$ you have to be careful factoring out (n+1)

Deniz Tuna Yalçın
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