Sequences series problem

Question

Show that square of any even natural number $(2n)^2$ is equal to sum of $n$ terms of some series of integers in A.P

Please don't solve the question . I want to have a general idea regarding these type of problems.

Answer 1

The general idea regarding these types of problems, is an inspection of the summation formula of $n$ terms of an arithmetic progression : $$ S_n = \frac{n}{2}(2a + (n-1)d) $$

where $a$ is the first term and $d$ is the common difference.

In your case, $S_n = 4n^2$ is specified, and $a$ and $d$ are to be decided.

Write this expression down and simplify it : $4n^2 = \frac n2 (2a + (n-1)d)$, so $8n = 2a + (n-1)d$.

This is true for all integers $n$, and the left hand side $8n$ is a polynomial.

To decide $a$ and $d$ , we now have the following way : assume that $a$ and $d$ depend polynomially on $n$ (that is, $a$ and $d$ are polynomials with the variable as $n$ e.g. $a = 2n^2 + 3n+8$, $d = 5$ etc.) Notice that the degree of the left hand side is $1$ ,so the degree of the right hand side is $1$. Hence, we may assume that $a$ must be of degree $1$ in $n$, and $d$ must be a constant, so that when we multiply it by $n-1$ we get at most a one degree polynomial.

So, write $a = bn+c$, and therefore : $8n = 2bn+2c+ dn - d$, so $(8-2b-d)n + (d-2c) = 0$.

When a polynomial is zero, so are all the coefficients. Hence, $2b+d = 8$ and $2c = d$.

At this stage, we have two equations in three variables. You will get a solution (if not, then the problem is impossible). For example, $d = 4$ gives you $b=c=2$.

Therefore, the sum of the first $n$ terms of the AP which starts with $2n+2$ and has common difference $4$, is $4n^2$.

For example, take $n = 3$ to get $8+12+16 = 36 = 4 \times 3^2$.

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There is more than one solution : $b = 0, d=8,c=4$ is another. Consequently, the sum of the first $n$ terms of an AP with first term $4$ and common difference $8$, is $4n^2$. For example, with $n = 4$ one gets $4 + 12 + 20 + 28 = 64 = 4 \times 4^2$.

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Summarizing, when $S_n$ is a polynomial in $n$, we assume that $a$ and $d$ are polynomials in $n$, such that $a$ has same degree as $S_n - 1$ (note : the $n$ on the RHS of the sum of AP formula comes into play here), and $d$ has same degree as $S_n - 2$, and then equate coefficients of RHS and LHS to get equations in the unknown coefficients of $a$ and $d$. Finding a solution of this then gives the desired result.

Answer 2

You may think about an arithmetic progression of integers as an integer-sided triangle with some additional appendices which ensures a perfect riemann sum.

The area of this triangle is the geometric illustration of the sum of this series added to both red and blue areas, which counts to $S=\frac{n*(nk)}{2}+\color{blue}{\frac{nk}{2}}+\color{red}{ln}$ considering $n$ the base, $nk$ the altitude of this triangle. and $l$ the the starting number of this series.

$S=\frac{n^2k+nk}{2}+ln=k(\frac{n^2+n}{2}+\frac{ln}{k})$

For this quantity to be four times a square. It requires $k$ to be $8u^2$ for any $u$ sufficiently large, and $\frac{ln}{k}=\frac{ln}{8u^2}$ to be $\frac{n+1}{2}$ which means $n=\frac{4u^2}{l-4u^2}$ for any arbitrary $l$ sufficiently small.