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I need to find a formula for $$ \sum_{i=1}^n (2i-1)^2 = 1^2 + 3^2 + \cdots + (2n-1)^2 $$ This problem is contained in Spivak's calculus ch2-2. I know that: $$ \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} $$ and $$ \sum_{i=1}^n i = \frac{n(n+1)}{2} $$ and I can use both these formulas.

Here is what I did: $$ \sum_{i=1}^n (2i-1)^2 = \sum_{i=1}^n (4i^2 - 4i + 1) = 4\sum_{i=1}^n i^2 - 4 \sum_{i=1}^n i + n $$ $$ \Rightarrow \sum_{i=1}^n (2i-1)^2 = 4 \frac{n(n+1)(2n+1)}{6} - 4 \frac{n(n+1)}{2} + n $$

Is that correct? Thanks in advance.

Eugene
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2 Answers2

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Note that $$k^2=\binom k2+\binom {k+1}2$$ Hence $$\begin{align} \sum_{i=1}^n(2i-1)^2&=\sum_{i=1}^n {2i-1\choose 2}+{2i\choose 2}\\ &=\sum_{i=1}^{2n}\binom i2\\ &={2n+1\choose 3}\qquad\blacksquare\\ \end{align}$$

NB - It is interesting to note that $$\begin{align} &\sum_{i=1}^n(2i-1)^2&&=1^2+3^3+5^2+\cdots+(2n-1)^2&&={2n+1\choose 3}\quad\text {as shown above, and}\\ &\sum_{i=1}^n (2i)^2&&=2^2+4^2+6^2+\cdots+(2n)^2&&={2n+2\choose 3}\\ \end{align}$$

Hypergeometricx
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One way to check is to take the (known) sum of the squares up to $2n$, then subtract $4$ times the sum of the squares up to halfway, because this removes the even squares: $$ \sum_{k=1}^n (2k-1)^2 = \sum_{k=1}^{2n} k^2 -\sum_{k=1}^n (2k)^2 = \frac{2n(2n+1)(4n+1)}{6} - 4\frac{n(n+1)(2n+1)}{6} = \frac{1}{3}n(2n+1)(4n+1-2(n+1)) = \frac{1}{3}n(2n+1)(2n-1) $$

Chappers
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