More Polynomial Equations
Find relationships between the polynomials a, b and c which are
polynomials in n giving the sums of the first n natural numbers,
squares and cubes respectively.
Age
16 to 18
Challenge level
Given any two polynomials in a single variable it is always possible to eliminate the variable and obtain a formula showing the relationship between the two polynomials.
$$\begin{eqnarray} a(n) &= &1 + 2 + 3 + ... + n \\ b(n) &= &1^2 + 2^2 + 3^2 + ... + n^2\\ c(n) &= &1^3 + 2^3 + 3^3 + ... + n^3. \end{eqnarray}$$
It is well known that $c(n) = a(n)^2$ . What are the relationships between $a(n)$ and $b(n)$ and between $b(n)$ and $c(n)$?
Here you need to start with $$\begin{eqnarray} a(n) &= &1 + 2 + 3 + ... + n &= & {1\over 2}n(n+1)\\ b(n) &= &1^2 + 2^2 + 3^2 + ... + n^2 &= & {1\over 6}n(n+1)(2n + 1)\\ c(n) &= &1^3 + 2^3 + 3^3 + ... + n^3 &= & {1\over 4}n^2(n+1)^2. \end{eqnarray}$$
These are well known results found in many text books and you will find proofs in some problems on this website. See Picture Story and Natural Sum.
Jim sent in this solution, using the ideas from our hints.
$a(n)=1+2+\ldots+n=\frac{n(n+1)}{2}$
$b(n)=1^2+2^2+\ldots+n^2=\frac{n(n+1)(2n+1)}{6}$
$c(n)=1^3+2^3+\ldots+n^3=\frac{n^2(n+1)^2}{4}$
It's obvious that $c=a^2$, from this.
Also, $2a=n^2+n$, so, solving the quadratic (and using the fact that $n> 0$), we get $n=\frac{-1+\sqrt{1+8a}}{2}$.
Now substitute this for $n$ in $b$, to get $b=\frac{n(n+1)(2n+1)}{6}=\frac{a}{3}\times(2n+1)=\frac{a}{3}\times\sqrt{1+ 8a}$. So $3b=a\sqrt{1+8a}$, so $9b^2=a^2+8a^3$.
Now we can combine these two expressions: $9b^2=c+8c\sqrt{c}$, so $8c\sqrt{c}= 9b^2-c$, so $64c^3=81b^4-18b^2c+c^2$.
(It's easy to check that the expressions above in terms of $n$ do work in this!)