Following Gauss's Path (II)

Algebra Level 3

It is said that, as a child, Gauss completed the tedious assignment of adding the first one hundred natural numbers in a few seconds by realizing that $1+100=2+99=\ldots=50+51$ , so the answer must be $50\times101=5050$ , since there are 50 such couples.

For the first $n$ natural numbers, the expression is: $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$

We know that the sum of the first $n$ cubes takes the form: $\sum_{k=1}^{n}k^3=A\,n^2+B\,n^3+C\,n^4$

Find the coefficients $A,B$ and $C$ . Express the solution as $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}$ .

Bonus : Factor the polynomial and compare it with $\frac{n(n+1)}{2}$ to find a beautiful identity that relates both sums!

The answer is 10.

1 solution

Gabriel Chacón
Nov 20, 2018

The coefficients $A,B$ and $C$ satisfy the following system of linear equations: $\begin{cases} n=1:\quad 1^2\,A\,+\,1^3\,B\,+\,1^4\,C\,=\,1^3\\ n=2:\quad 2^2\,A\,+\,2^3\,B\,+\,2^4\,C\,=\,1^3+2^3\\ n=3:\quad 3^2\,A\,+\,3^3\,B\,+\,3^4\,C\,=\,1^3+2^3+3^3 \end{cases}$ The solution is: $A=\dfrac{1}{4},B=\dfrac{1}{2},C=\dfrac{1}{4}$ . Expressed in the form demanded, the answer is $\boxed{\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=10}$ .

Bonus : The solution polynomial $\frac{1}{4}\,n^2+\frac{1}{2}\,n^3+\frac{1}{4}\,n^4$ can be easily factored as $\frac{n^2(n+1)^2}{4}$ , which is the square of the sum of the the first $n$ natural numbers. The identity referred to is then: $\boxed{1^3+2^3+\ldots+n^3=(1+2+\ldots+n)^2}$ Isn't it beautiful!

But how does one know that the formula $\displaystyle \sum_{k=1}^{n}k^3=A\,n^2+B\,n^3+C\,n^4$ work for all positive integer $n$ ?

Pi Han Goh - 2 years, 6 months ago

Well, I don't know. I need your support here.

Gabriel Chacón - 2 years, 6 months ago

Use finite difference to show that $\displaystyle \sum_{k=1}^{n}k^p$ is a polynomial of $n$ in degree $p+1$ .

Pi Han Goh - 2 years, 6 months ago

Following Gauss's Path (II)

The answer is 10.

1 solution

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