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I am trying to find the nth partial sum of this series: $S(n) = 2(n+1)^2$

I found the answer on WolframAlpha:

$\sum_{n=0}^m (1+2n)^2 =\frac{1}{3}(m+1)(2m+1)(2m+3)$

How can I calculate that sum, without any software?

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  • $\begingroup$ It's a sum of squares. You may learn from math.stackexchange.com/q/188602/290189 $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Oct 1 '18 at 20:41
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    $\begingroup$ $2(n+1)^2\ne(1+2n)^2$. Which one are you trying to calculate? $\endgroup$ – Andrei Oct 1 '18 at 20:43
  • $\begingroup$ Hint: \begin{eqnarray*} \sum_{n=0}^{m} n^2=\frac{m(m+1)(2m+1)}{6} \end{eqnarray*} $\endgroup$ – Donald Splutterwit Oct 1 '18 at 20:44
  • $\begingroup$ @Andrei Sorry, that is a mistake. I am trying to calculate $(1+2n)^2$ $\endgroup$ – GKEdv Oct 1 '18 at 20:48
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$$S(n)=(1+2n)^2=1+4n+4n^2$$ You can now use the following $$\sum_{n=0}^m1=m+1\\\sum_{n=0}^mn=\frac{m(m+1)}{2}\\\sum_{n=0}^mn^2=\frac{m(m+1)(2m+1)}{6}$$

Alternatively, compute the first 4-5 elements. The sum of a polynomial of order $p$ will be a polynomial of order $p+1$ in the number of terms. Find the coefficients, then prove by induction

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  • $\begingroup$ Thanks, the first approach is very easy to understand. Can you elaborate on the alternative approach or point me to more information, example on that? $\endgroup$ – GKEdv Oct 1 '18 at 21:20
  • $\begingroup$ If you look at en.wikipedia.org/wiki/Faulhaber%27s_formula#Summae_Potestatum you can get that the sum of $n^p$ is a polynomial of order $p+1$. Adding then the rest of the smaller powers will change the coefficients, but not the order of the polynomial. $\endgroup$ – Andrei Oct 1 '18 at 21:44
  • $\begingroup$ The first sum should be $m\color{red}{+1}$ ... it starts at zero! $\endgroup$ – Donald Splutterwit Oct 2 '18 at 1:39
  • $\begingroup$ You are right. It does not matter for the other ones. Thanks. I'll fix it $\endgroup$ – Andrei Oct 2 '18 at 3:14
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$\sum_\limits{i=0}^n 2(i + 1)^2 = 2\sum_\limits{i=1}^{n+1} i^2$

Which gets to the meat of the question, what is $\sum_\limits{i=1}^n i^2$?

There are a few ways to do this. I think that this one is intuitive. enter image description here

In the first triangle, the sum of $i^{th}$ row equals $i^2$

The next two triangles are identical to the first but rotated 120 degrees in each direction.

Adding corresponding entries we get a triangle with $2n+1$ in every entry. What is the $n^{th}$ triangular number?

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

To find: $\sum_\limits{i=1}^{n+1} i^2 $, sub $n+1$ in for $n$ in the formula above.

$\sum_\limits{i=0}^n 2(i + 1)^2 = \frac {(n+1)(n+2)(2n+3)}{3}$

Another approach is to assume that $S_n$ can be expressed as a degree $3$ polynomial. This should seem plausible

$S(n) = a_0 + a_1 n + a_2 n^2 + a_3n^3\\ S(n+1) = S(n) + 2(n+2)^2\\ S(n+1) - S_n = 2(n+2)^2\\ S(n+1) = a_0 + a_1 (n+1) + a_2 (n+1)^2 + a_3(n+1)^3\\ a_0 + a_1 n+a_1 + a_2 n^2 + 2a_2n+a_21 + a_3n^3 + 3a_3n^2 + 3a_3n + 1\\ S(n+1) - S(n) = (a_1 + a_2 + a_3) + (2a_2 + 3a_3) n + 3a_3 n^2 = 2n^2 + 4n + 2$

giving a system of equations:

$a_1 + a_2 + a_3 = 2\\ 2a_2 + 3a_3 = 4\\ 3a_3 = 1\\ a_0 = S(0)$

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