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Statement: $1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = (n/3)*(2n-1)*(2n+1)$

Proof by induction

-Base case: when $n = 1$

$1^2 = 1/3 * (2 * 1 -1) * (2 * 1 +1) = 1$

$1=1$ hence statement holds for $n = 1$

-Inductive step

assume $n = k$ is true

then $1^2 + 3^2 + 5^2 + .. + (2k-1)^2 = (k/3) * (2k - 1) * (2k + 1)$

let $n = k+1$

then $1^2 + 3^2 + 5^2 + ... (2k - 1)^2 + (2(k+1)-1)^2$

it follows that $1^2 + 3^2 +5^2 .. + (2k-1)^2 = (k/3) * (2k-1) * (2k+1)$

such that $(k/3) * (2k-1)*(2k+1) + (2(k+1) -1)^2 = ((k+1)/3) * (2(k+1) - 1) * (2(k+1) +1)$

hence: $(4k^3+12k^2 + 11k + 3)/3 = (4k^3+12k^2 + 11k + 3)/3$

thus the statement holds for $n+1$ and the statement is therefore true

Is this proof correct? Am I missing anything or detail that I can add?

Marc van Leeuwen
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jn025
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  • "it follows that..." what do you mean? The inductionhypothese is a statement ahead (and does not need to be repeated). It is not something that "follows". – drhab Sep 20 '14 at 14:28
  • I think your induction step isn't worded quite right. You don't just want to assume $n=k$ is true. You want to assume there is some $n$ such that for your proposition holds for all $k$ where $1 \leq k \leq n$. Then you can explore the cases of $n+1$ – graydad Sep 20 '14 at 14:38
  • See also. – Lucian Sep 20 '14 at 15:17

2 Answers2

1

After mentioning the case when $n=1$, the way you are writing looks strange to me.

Also, note that you don't need to expand them.

Assuming that $1^2+3^2+\cdots +(2k-1)^2=\frac{k(2k-1)(2k+1)}{3}$ holds, we have $$\begin{align}1^2+3^2+\cdots +(2k-1)^2+(2(k+1)-1)^2&=\frac{k(2k-1)(2k+1)}{3}+(2k+1)^2\\&=\frac{2k+1}{3}\left(k(2k-1)+3(2k+1)\right)\\&=\frac{2k+1}{3}(2k^2+5k+3)\\&=\frac{2k+1}{3}(k+1)(2k+3)\\&=\frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}.\end{align}$$

mathlove
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1

Your line "hence something complicated is equal to itself" is unconvincing and suggests that you might do better with a different order, perhaps like $$1^2 + 3^2 + 5^2 + ... (2k - 1)^2 + (2(k+1)-1)^2$$ $$=(k/3) * (2k-1)*(2k+1) + (2(k+1) -1)^2$$ $$=(4k^3+12k^2 + 11k + 3)/3$$ $$= ((k+1)/3) * (2(k+1) - 1) * (2(k+1) +1)$$ $$= (n/3)*(2n-1)*(2n+1) \text{ when } n=k+1$$

Henry
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