When a Rectangle and Cube are Summed!

Algebra Level 3

Let $f(n) = \dfrac{g(n)}{h(n)}$ , where $g(n) = 1^3 + 2^3 + \cdots + n^3$ and $h(n) = 2 + 6 + 12 + 20 + 30 + \cdots + n(n+1)$ .

If the value of $f(100)$ can be expressed as $\dfrac mn$ , where $m$ and $n$ are coprime positive integers, find $m+n$ .

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The answer is 2559.

1 solution

Fidel Simanjuntak
Feb 9, 2017

We notice that $f(n) = \dfrac{ \dfrac{1}{4} n^2(n+1)^2}{ \dfrac{1}{3} n(n+1)(n+2) }$ , for which the numerator is the formula for the sum of the first $n^{th}$ term in cubic number sequence $\left(1,8,27,64, \cdots \right)$ , and the demominator is the formula for the sum of the first $n^{th}$ term in rectangle number sequence $\left( 2, 6, 12, 20, 30, \cdots \right)$ .

Now, $f(100) = \dfrac{ \dfrac{1}{4} \cdot 100^2 \cdot 101^2 }{ \dfrac{1}{3} \cdot 100 \cdot 101 \cdot 102 }$ .

Simplify, we have $f(100) = \dfrac{2525}{34} = \dfrac{m}{n}$ .

Hence, our final answer $m+n = 2559$ .

Note: The reason why I titled this problem with "When a Rectangle and Cube are Summed!" is because the things that I mentioned above. First, the " hint " says "Summed", so at least we have to find the formula of the sum of the first $n^{th}$ term in cubic number sequence, and rectangle number sequence as well. And do some trial and error to clarify the fraction with those formula that we have. And by the information that we have, we can easily get that $f(n) = \dfrac{ \dfrac{1}{4} n^2(n+1)^2}{ \dfrac{1}{3} n(n+1)(n+2) }$

When a Rectangle and Cube are Summed!

Try another problem on my set! Let's Practice

The answer is 2559.

1 solution

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