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I need to know if I am doing this right. I have to prove that $1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = \frac{n(2n-1)(2n+1))}{3}$

So first I did the base case which would be $1$.

$1^2 = (1(2(1)-1)(2(1)+1)) / 3 1 = 3/3 1 = 1$ Which is right.

Then I assumed true for k so $1^2 + 3^2 + 5^2 + ... + ((2k-1)^2 = k(2k-1)(2k+1)) / 3$

This is where I get lost. I know that the next step is to prove its true for k+1, but I am lost on what to do here. Do I actually simplify down what I assumed true for k? Am I right so far?

Martin Sleziak
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Jebus Crust
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    You are right so far.Write down the equation for the case $k$, which you have to prove, and see how you can gert there. As a hint, add $(2k+1)^2$ to both sides, so that the left hand side becomes your L.H.S for the case $k+1$, and then see if you can simplify the right hand side into the R.H.S of the case $k+1$, if necessary by expanding the brackets completely, simplifying and regrouping. In this case it seems easy, if you like I'll elaborate further. – Sarvesh Ravichandran Iyer Mar 30 '16 at 03:05
  • See also this question and perhaps also other posts linked there. – Martin Sleziak Mar 31 '16 at 15:57

3 Answers3

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Let your statment be $A(n)$. You want to show it holds for all $n \in \mathbb{N}$. You use the principle of induction to establish a chain of implications starting at $A(1)$ (you did that one). What is left to show is $$ A(n) \Rightarrow A(n+1) $$ This means you consider $n$ fixed and try to proof $A(n+1)$. For this you can assume that $A(n)$ is true, or in another variant of the principle of induction: $A(1), \dotsc, A(n)$ are true.

You have: \begin{align} A(n) &: \sum_{k=1}^n (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3} \\ A(n+1) &: \sum_{k=1}^{n+1} (2k-1)^2 = \frac{(n+1)(2(n+1)-1)(2(n+1)+1)}{3} \\ \end{align} One way is to split $$ \sum_{k=1}^{n+1} (2k-1)^2 = \left(\sum_{k=1}^{n} (2k-1)^2 \right) + (2(n+1)-1)^2 $$ and to use $A(n)$.

mvw
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Hypothesis: $$\sum \limits_{k=1}^n (2k-1)^2 = \dfrac{n(2n-1)(2n+1)}{3} $$

For $n+1$: $$\sum \limits_{k=1}^{n+1} (2k-1)^2 = \dfrac{(n+1)(2n+1)(2n+3)}{3} = \dfrac{4n^3 - n}{3} + (2n+1)^2 = \dfrac{n(2n-1)(2n+1)}{3} + (2n+1)^2$$

which is exactly what we wanted.

Airdish
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This is not a proof by induction, but is included for information of those who might be interested.

Note that $$r^2=\binom r2+\binom {r+1}2$$ Hence $$\begin{align} &\color{blue}{1^2}+\color{purple}{3^2}+\color{green}{5^2}+\cdots+\color{red}{(2n-1)^2}\\ &=\color{blue}{\binom 12+\binom 22}+ \color{purple}{\binom 32+\binom 42}+ \color{green}{\binom 52+\binom 62}+\cdots + \color{red}{\binom {2n-1}2+\binom {2n}2}\\ &=\sum_{r=1}^{2n}\binom r2\\ &=\binom {2n+1}3 =\frac {n(2n-1)(2n+1)}3\qquad\blacksquare\end{align}$$

Hypergeometricx
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