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In Spivak's Calculus, Chapter 2, the second problem of the set asks you to find a formula for the following series:

$$\sum_{i=1}^n(2i-1)$$ and $$\sum_{i=1}^n(2i-1)^2$$ Now, for the former, this was fairly straightforward: it sums to $i^2$ and I have a proof that I'm happy with. However, for the latter, I cannot identify a pattern in terms of $n$ to begin building a proof. The series proceeds $1^2 + 3^2 + 5^2 + 7^2 + \cdots + (2n-1)^2$, so the sum for the first few indices would be $1, 10, 35,$ and $84.$ Not only have I failed to come up with an expression of this in terms of $n$, but I'm not even sure what I should be considering to lead me to such a formula.

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marked as duplicate by Hans Lundmark, Claude Leibovici, ccorn, cansomeonehelpmeout, Christopher Jun 8 '18 at 12:16

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  • $\begingroup$ Are you also supposed to give a direct proof by induction, or can you use other known formulæ like the sum of the first $2n$ squares? $\endgroup$ – Bernard Jun 6 '18 at 20:33
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Hint : Expand the summand & use \begin{eqnarray*} \sum_{i=1}^{n} i^2 =\frac{n(n+1)(2n+1)}{6}. \end{eqnarray*} You should get to \begin{eqnarray*} \sum_{i=1}^{n} (2i-1)^2 =\frac{n(4n^2-1)}{3}. \end{eqnarray*}

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  • $\begingroup$ Thanks to a combination of your hint and dezdichado's below, I managed to figure it out. Thank you very much! $\endgroup$ – user242007 Jun 6 '18 at 21:44
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If you know what $1^2+2^2+...+n^2=f(n)$ sums up to, then your desired sum is simply: $$1^2+3^2+...+(2n-1)^2 = f(2n) - (2^2+4^2+...+(2n)^2) = f(2n) - 4f(n).$$

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    $\begingroup$ Your comment helped me become more aware of incorporating existing known formulae in these exercises. Between that and Donald's comment above, I was able to solve the problem. Thank you. $\endgroup$ – user242007 Jun 6 '18 at 21:44
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$$\sum_{i=1}^n (2i-1)^2$$

$$=\sum_{i=1}^n 4i^2-4i+1$$

$$=4\sum_{i=1}^n i^2-4\sum_{i=1}^n i+\sum_{i=1}^n1$$

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  • $\begingroup$ Could you elaborate on this? In trying to implement it with an example value of i = 4, I end up with 81, when the actual value in the series appears to be 84. $\endgroup$ – user242007 Jun 6 '18 at 21:26
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$\sum_\limits{k=1}^n (2k - 1)^2 = \sum_\limits{k=1}^n 4k^2 - \sum_\limits{k=1}^n 4k + \sum_\limits{k=1}^n 1 = 4\left(\sum_\limits{k=1}^n k^2\right) - 2n(n+1) + n$

You need a way to find $\sum_\limits{k=1}^n k^2$

There are lots of other ways to derive this... but this one is nice.

Write the numbers in a triangle

$\begin {array}{} &&&1\\&&2&&2\\&3&&3&&3\\&&&\vdots&&\ddots\\n&&n&\cdots\end{array}$

The sum of each row is $k^2.$ And, the sum of the whole trianglular array is $\sum_\limits{k=1}^n k^2$

Take this triangle and rotate it 60 degrees clockwise and 60 degrees counter clockwise and sum the 3 triangles together.

$\begin {array}{} &&&1\\&&2&&2\\&3&&3&&3\end{array}+\begin {array}{} &&&n\\&&n-1&&n\\&n-2&&n-1&&n\\\end{array}+\begin {array}{} &&&n\\&&n&&n-1\\&n&&n-1&&n-2\end{array}$

And we get:

$\begin {array}{} &&&2n+1\\&&2n+1&&2n+1\\&2n+1&&2n+1&&2n+1\\&&&\vdots&&\ddots\end{array}$

What remains how many elements are in the triangular array? $\sum_\limits{k=1}^n k$

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

Another way to do it is:

$\sum_\limits{k=1}^n (k+1)^3 - k^3 = (n+1)^3-1\\ \sum_\limits{k=1}^n (3k^2 + 3k + 1)\\ \sum_\limits{k=1}^n 3k^2 + \sum_\limits{k=1}^n 3k + \sum_\limits{k=1}^n 1\\ \sum_\limits{k=1}^n 3k^2 + 3\frac {n(n+1)}{2} + n = n^3 + 3n^2 + 3n \\ 3\sum_\limits{k=1}^n k^2 = n^3 + \frac 32 n^2 + \frac 12 n\\ \sum_\limits{k=1}^n k^2 = \frac 16 (n)(n+1)(2n+1)$

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  • $\begingroup$ I like the rotation method, that works out very nicely! $\endgroup$ – abiessu Jun 6 '18 at 20:59
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    $\begingroup$ It is nice in that it is intuitive, but the other method can be generalized. $\endgroup$ – Doug M Jun 6 '18 at 21:01
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You can do it by telescoping. Given a sequence $f(n)$ define the sequence $\Delta f(n)=f(n+1)-f(n)$ and put $(n)_m=n(n-1)\dotsb(n-m+1)$. One can check that $\Delta (n)_m=m(n)_{m-1}$ for $m\geq 1$. Now note that $$ (2i-1)^2=4i^2-4i+1=4[(i)_2+i] -4i+1=4(i)_2+1$$ whence $$ \Delta\left(\frac{4}{3}(i)_3+i\right)=(2i-1)^2. $$ In particular $$ \sum_{i=1}^n (2i-1)^2=\sum_{i=1}^n \Delta\left(\frac{4}{3}(i)_3+i\right)=\left[\frac{4}{3}(i)_3+i\right]_{i=1}^{n+1} $$ with the notation $[g(i)]_{i=a}^b=g(b)-g(a)$.

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