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Is there an analytical expression for the summation $$1^2+3^2+5^2+\cdots+(2n-1)^2,$$ and how do you derive it?

ViHdzP
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LWZ
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5 Answers5

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$$ 1^2+3^2+5^2+\cdots+(2n-1)^2 = \sum_{i=1}^{n}(2i-1)^2 = \sum_{i=1}^{n}(4i^2)- \sum_{i=1}^{n}(4i) + \sum_{1}^{n}1=\dots.$$

Now, you need the identities

$$ \sum_{i=1}^{n}1 = n, $$

$$ \sum_{i=1}^{n}i = \frac{n(n+1)}{2}, $$

$$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}.$$

Community
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Mhenni Benghorbal
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The following is a very general technique. In particular, it's probably not the best way to solve this sort of problem in the beginning. But it is worth knowing this technique in the long run.

We use the result that:$$\sum_{k=0}^{n-1} \binom{k}{i} = \binom{n}{i+1}$$

Write: $$(2k+1)^2=8\binom{k}{2} +8\binom{k}{1}+ \binom{k}{0}$$ Then $$\begin{align}\sum_{k=0}^{n-1} (2k+1)^2 &=\sum_{k=0}^{n-1}\left( 8\binom{k}{2} + 8\binom{k}{1} + \binom{k}{0}\right)\\&=8\binom{n}{3}+8\binom{n}{2} + \binom{n}{1}\end{align}$$

This method can be used to solve $\sum_{k=0}^{n-1} p(k)$ for any polynomial $p$. The result is always a polynomial of one degree higher than the degree of $p$.

Thomas Andrews
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  • I'm sorry, could you please explain a little more what does this bracket notation mean? – LWZ Jul 07 '13 at 03:43
  • Ah, if you don't know the "binomial coefficient," then this will seem mysterious. See: http://en.wikipedia.org/wiki/Binomial_coefficient – Thomas Andrews Jul 07 '13 at 03:44
  • @ThomasAndrews Do you have a link (without induction) of the first result? – Ian Mateus Jul 07 '13 at 03:45
  • @IanMateus That's a pretty common result, but I'm not sure the best place for it. The right hand side counts subsets of size $i+1$ of ${1,2,\dots,n}$, while the the $k$th term counts the same subsets where the largest element is $k+1$. It's also pretty easy to prove via induction on $n$... – Thomas Andrews Jul 07 '13 at 03:48
  • Found it on Wikipedia. Thanks! – Ian Mateus Jul 07 '13 at 03:52
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One can give a combinatorial version of the argument below. But it takes some time to tell the right story. So for now we settle for the magic math approach. We have $$\binom{2k+1}{3}-\binom{2k-1}{3}=\frac{(2k+1)(2k)(2k-1)-(2k-1)(2k-2)(2k-3)}{3!}.\tag{1}$$ Note that the above equation even holds when $k=1$, if we use the convention that $\binom{1}{3}=0$.

The right-hand side of (1) magically simplifies to $(2k-1)^2$. It follows that our sum $1^2+3^2+5^2+\cdots +(2n-1)^2$ is equal to $$ \binom{3}{3}-\binom{1}{3} + \binom{5}{3}-\binom{3}{3} + \binom{7}{3}-\binom{5}{3} + \cdots + \binom{2n+1}{3}-\binom{2n-1}{3} .$$ Note the wholesale cancellation. We get $$1^2+3^2+5^2+\cdots +(2n-1)^2=\binom{2n+1}{3}.$$

André Nicolas
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    Kind of working backwards, but still nice. I suppose this would be "obvious" if we realized that $k^2=\binom{k}{2}+\binom{k+1}{2}$. – Thomas Andrews Jul 07 '13 at 04:01
  • There are various combinatorial ways to show that the sum is indeed $\binom{2n+1}{3}$. Here is a cute thing you might like. The formula for the sum of the first $n$ squares is kind of strange, with that weird $2n+1$. Actually, it is $n$ and $n+1$ that are weird. One can use a combinatorial argument to show that the sum is $\frac{1}{4}\binom{2n+2}{3}$. – André Nicolas Jul 07 '13 at 04:12
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Hint: $1^2+2^2+3^2+...+k^2=k(k+1)(2k+1)/6$.

JSCB
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1² + 3² + 5² + ...

Tn = [1 + (n – 1)(2)]² = (2n – 1)² = 4n² – 4n + 1

Total sum = ∑4n² – ∑4n + n

= 4n(n + 1)(2n+1)/6 – 4n(n + 1)/2 + n

= 2n(n + 1)(2n+1)/3 – 2n(n + 1) + n

= 2n[(n + 1)(2n+1)/3 – 1] + n

= 2n[(n + 1)(2n + 1 – 3/3] + n

= 2n[(n + 1)(2n – 2/3] + n

= 4n[(n + 1)(n – 1/3] + n

= 4n[(n + 1)(n – 1 + 3n]/3

= n(2n + 1)(2n – 1)/3

Sunil Pradhan
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