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$S_{2n}+4S_{n}=n(2n+1)^2$, where $S_{2n}$ is the Sum of the squares of the first $2n$ natural numbers, $S_{n}$ is the Sum of the squares of the first $n$ natural numbers.

when, $n=2$

$S_{2n}=S_{4}=1^2+2^2+3^2+4^2=30$

$S_{n}=S_{2}=1^2+2^2=5$

$S_{4}+4S_{2}=2(2*2+1)^2=50$

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    thanks for the comments. i will pay heed to the comments.2012-09-14

6 Answers 6

5

Here is a closed form solution to your recurrence relation obtained by Maple,

$ s(n)={n}^{2}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}}s \left( 1 \right) +\frac{{n}^{3}}{3}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}} \left( \left( -1 \right)^{{\frac {\ln \left( n \right) }{\ln \left( 2 \right) }}} \right)^{-1}+\frac{{n}^{2}}{2}{n}^{{\frac {i\pi }{ \ln \left( 2 \right) }}} \left( \left( -1 \right) ^{{\frac {\ln \left( n \right) }{\ln \left( 2 \right) }}} \right) ^{-1}+\frac{1}{6}\,n{n}^ {{\frac {i\pi }{\ln \left( 2 \right) }}} \left( \left( -1 \right) ^{ {\frac {\ln \left( n \right) }{\ln \left( 2 \right) }}} \right) ^{-1 }-{n}^{2}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}} \,$

Here is a more compact form

$ s(n) = \left( {n}^{2}\cos \left( {\frac {\pi \,\ln \left( n \right) }{\ln \left( 2 \right) }} \right) +i{n}^{2}\sin \left( {\frac {\pi \,\ln \left( n \right) }{\ln \left( 2 \right) }} \right) \right) s \left( 1 \right) -{n}^{2}\cos \left( {\frac {\pi \,\ln \left( n \right) }{\ln \left( 2 \right) }} \right) +\frac{{n}^{3}}{3}+\frac{{n}^{2}}{2} +\frac{n}{6}-i{n}^{2}\sin \left( {\frac {\pi \,\ln \left( n \right) }{\ln \left( 2 \right) }} \right) \,.$

where $s(1)$ is your initial condition. If you plug in $s(1)=1$ in the above formula you get the simple formula, just as it has been mentioned in the comments,

$ \frac{n}{6} \left( n+1 \right) \left( 2\,n+1 \right) \,,$

which is equal to $ \sum_{i=1}^{n} i^2 $.

Note

If you are interested only in finding sums of the form $ \sum_{i=1}^{n} i^m \,, m=1,2,3,\dots $, then they are simple techniques to find them. See here.

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    @alex.jordan: I had the same kind of puzzlement, but maybe Maple was looking for a solution given by a single expression, and valid for all real (or complex) values of $n$?2012-09-01
4

Let $S_n=an^3+bn^2+cn+d$ where $a,b,c,d$ are rational numbers.

So, $S_{2n}+4S_n=n^3 12a+n^2 8b + n 6c+5d$

$\implies n^3 12a+n^2 8b + n 6c+5d= n(2n+1)^2=4n^3+4n^2+n$

Comparing the coefficients of the different powers on $n$,

$12a=4,8b=4,6c=1,d=0$

So, $6S_n=2n^3+3n^2+n=n(n+1)(2n+1)\implies S_n=\frac{n(n+1)(2n+1)}{6}$

Also, $S_n -S_{n-1}=n^2\implies S_n=n^2+S_{n-1}=\sum_{1≤r≤n}r^2+S_0=\sum_{1≤r≤n}r^2$

Observe that, we don't need to know the nature or formula of $S_n$. Solution of any such difference equation of any positive integer degree can be attempted this way.

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    @Tunococ It is. See my comment on alex.jordan's answer.2012-08-31
2

With the additional information you provided about $S_n$ (the sum of squares of the first $n$ integers), there's a neat solution (among other solutions) that uses the perturbation method described in Concrete Mathematics.

Let $C_n$ denote the sum of cubes of the first $n$ natural numbers. Then

\begin{equation} \begin{split} C_{n+1} =& C_n + (n+1)^3 = \sum_{k=1}^{n+1}k^3 = \sum_{k=0}^{n}(k+1)^3 = \sum_{k=0}^{n} k^3 + 3k^2 + 3k + 1 \\ =&\sum_{k=0}^{n}k^3 + 3\sum_{k=0}^{n}k^2 + 3\sum_{k=0}^{n}k + \sum_{k=0}^{n}1 = C_n + 3S_n + 3\dfrac{n(n+1)}{2} + (n+1). \end{split} \end{equation}

Hence \begin{equation} \begin{split} S_n =& \dfrac{(n+1)^3}{3} - \dfrac{n(n+1)}{2} - \dfrac{n+1}{3} = \dfrac{(n+1)(2(n+1)^2 - 3n - 2)}{6} \\ =& \dfrac{(n+1)(2n^2+n)}{6} = \dfrac{n(n+1)(2n+1)}{6}. \end{split} \end{equation}

2

$S_{2n}+4S_{n}=n(2n+1)^2$

$S_{2n}=S_{n}+S_{(n+1,2n)}$ -------(A)

$S_{(n+1,2n)}=(n+1)^2+(n+2)^2+(n+3)^2+\cdots+(2n)^2$

$S_{(n+1,2n)}=(n+1)^2+(n+2)^2+(n+3)^2+\cdots+(n+n)^2$

$S_{(n+1,2n)}=n(n)^2+(1^2+2^2+3^2+\cdots+n^2)+(2n)(1+2+3+\cdots+n)$

$S_{(n+1,2n)}=n^3+S_{n}+(2n)\frac{n(n+1)}{2}$

$S_{(n+1,2n)}=n^3+S_{n}+n^2(n+1)$

$S_{(n+1,2n)}=n^3+S_{n}+n^3+n^2$

$S_{(n+1,2n)}=S_{n}+2n^3+n^2$ -------(B)

we, have from (A) and (B),

$S_{2n}=S_{n}+S_{n}+2n^3+n^2$

$S_{2n}=2S_{n}+2n^3+n^2$

we, now have,

$6S_{n}+2n^3+n^2=n(2n+1)^2$

$6S_{n}+2n^3+n^2=n(4n^2+4n+1)$

$6S_{n}+2n^3+n^2=4n^3+4n^2+n$

$6S_{n}=2n^3+3n^2+n$

$6S_{n}=n(n+1)(2n+1)$

$S_{n}=\frac{n(n+1)(2n+1)}{6}$

1

You could use induction.

Assuming that $S_{2n}+4S_n=n(2n+1)^2$ then add terms to both sides so that the left side increments its index: $ \begin{align} &S_{2n}+4S_n+(2n+1)^2+(2n+2)^2+4(n+1)^2\\ &=n(2n+1)^2+(2n+1)^2+(2n+2)^2+4(n+1)^2\\ S_{2(n+1)}+4S_{n+1}&=n(2n+1)^2+(2n+1)^2+(2n+2)^2+4(n+1)^2\\ &=(n+1)(2n+1)^2+4(n+1)^2+4(n+1)^2\\ &=(n+1)(2n+1)^2+8(n+1)^2\\ &=(n+1)[(2n+1)^2+8(n+1)]\\ &=(n+1)[4n^2+4n+1+8n+8]\\ &=(n+1)[4n^2+12n+9]\\ &=(n+1)(2n+3)^2\\ &=(n+1)(2(n+1)+1)^2\\ \end{align}$

The base case is established in your question.

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    You are absolutely right. In fact the question the OP wants to be solved seems to be as follows. Assume that (1.) $S_n=1^2+2^2+\cdots+n^2$ for every $n$ and that (2.) $S_{2n}+4S_n=n(2n+1)^2$ for every $n$, show that (3.) $S_n=\frac16n(n+1)(2n+1)$ for every $n$. (Of course, (1.) implies (2.) hence all this is rather odd but...)2012-08-31
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Hint $\rm\quad S_n =\, \sum c_k n^k\ \Rightarrow\ S_{2n} + 4\, S_n\, =\: \sum\ (2^k\!+\!4)\ c_k\ =\ 4\, n^3+4\,n^2 + n,\:$ therefore

$\rm S_n\, =\ \frac{4}{2^{\color{#C00}3}\!+\!4} n^\color{#C00}3 +\frac{4}{2^\color{#0A0}2\!+\!4}n^{\color{#0A0}2} + \frac{1}{2^\color{brown}1\!+\!4} n^{\color{brown}1}\ =\ \frac{n^3}3+\frac{n^3}2 + \frac{n}6\ =\ \frac{n(n+1)(2n+1)}6$