Crazy Cubes...and Fractions

Before I start with the problem, I suggest we should all remember the $3$ summation equations. They are:

$1 + 2 + 3 + ..... + n = \frac{n(n + 1)}{2}$

$1^{2} + 2^{2} + 3^{2} + ..... + n^{2} = \frac{n(n + 1)(2n+1)}{6}$

$1^{3} + 2^{3} + 3^{3} + ..... + n^{3} = \frac{n^{2}(n + 1)^{2}}{4}$

Now, coming to our problem, we have

$\frac{1^{3} + 3^{3} + 5^{3} + ..... + (2m -1)^{3}}{2^{3} + 4^{3} + 6^{3} + ..... + (2m)^{3}} = \frac{721}{800}$

The numerator can be rewritten as,

$\frac{(1^{3} + 2^{3} + 3^{3} + 4^{3} + ..... + (2m)^{3}) - (2^{3} + 4^{3} + 6^{3} + 8^{3} + ..... + (2m)^{3})}{2^{3} + 4^{3} + 6^{3} + ..... + (2m)^{3}} = \frac{721}{800}$

$\frac{(1^{3} + 2^{3} + 3^{3} + 4^{3} + ..... + (2m)^{3})}{2^{3} + 4^{3} + 6^{3} + ..... + (2m)^{3}} - \frac{(2^{3} + 4^{3} + 6^{3} + 8^{3} + ..... + (2m)^{3})}{2^{3} + 4^{3} + 6^{3} + ..... + (2m)^{3}} = \frac{721}{800}$

$\frac{(1^{3} + 2^{3} + 3^{3} + 4^{3} + ..... + (2m)^{3})}{2^{3} + 4^{3} + 6^{3} + ..... + (2m)^{3}} - 1 = \frac{721}{800}$

$\frac{(1^{3} + 2^{3} + 3^{3} + 4^{3} + ..... + (2m)^{3})}{2^{3}(1^{3} + 2^{3} + 3^{3} + ..... + (m)^{3})} - 1 = \frac{721}{800}$

Apply the formula of sum of cubes, we have

$\frac{\frac{(2m)^{2}(2m + 1)^{2}}{4}}{2^{3}\frac{(m)^{2}(m + 1)^{2}}{4}} - 1 = \frac{721}{800}$

$\frac{(2m)^{2}(2m + 1)^{2}}{2^{3}(m)^{2}(m + 1)^{2}} - 1 = \frac{721}{800}$

$\frac{(2m + 1)^{2}}{2(m + 1)^{2}} - 1 = \frac{721}{800}$

$\frac{(2m + 1)^{2}}{2(m + 1)^{2}} = \frac{721}{800} + 1$

$\frac{(2m + 1)^{2}}{2(m + 1)^{2}} = \frac{1521}{800}$

$\frac{(2m + 1)^{2}}{(m + 1)^{2}} = \frac{1521}{400}$

Taking square root on both sides,

$\frac{2m + 1}{m + 1} = \frac{39}{20}$

Cross-multiply,

$40m + 20 = 39m + 39$

$m = 19$

That's the answer!

The formula for the sum of the first $n$ perfect cubes is $(\frac{n(n+1)}{2})^2$ .

$\frac{1^3+3^3+...+(2m-1)^3}{2^3+4^3+...+(2m)^3}=\frac{(1^3+2^3+3^3+...+(2m)^3)-(2^3+4^3+...(2m)^3)}{2^3(1^3+2^3+3^3+...+m^3)}=\frac{((2m)(2m+1)/2)^2-8((m)(m+1)/2)^2}{8((m)(m+1)/2)^2}=\frac{(4m^2+2m)^2-8(m^2+m)^2)}{8(m^2+m)^2}=\frac{721}{800}$

$\frac{(4m+2)^2-8(m+1)^2)}{8(m+1)^2}=\frac{721}{800}$

Simplifying the expression, then cross-multiplying, then moving all the terms to the LHS yields a quadratic equation. Solving the equation gets m= $\boxed{19}$

First, we must know the formula for the sum of cubes. The sum of a sequence of cubes $1^3 + 2^3 + 3^3 + ... + m^3$ is equal to $\left(\frac{m(m+1)}{2}\right)^2$ . We need to reduce both sides of the fraction to such a sequence. The denominator of the fraction, $2^3 + 4^3 + 6^3 + ... + (2m)^3$ , can be written out as the sum of cubes by taking out the common factor of $2^3 = 8$ , so the value is $8(1^3 + 2^3 + 3^3 + ... + m^3)$ . Using our formula for the sum of cubes, the simplified value of the denominator is $8\left(\frac{m(m+1)}{2}\right)^2$ .

Now for the numerator, which is slightly harder. It can be expressed as the sum of the cubes from $1$ to $2m$ minus the sum of even cubes from $1$ to $2m$ (the denominator), which is eight times the sum of cubes from $1$ to $m$ . If we simplify using our formula for the sum of cubes, we get $\left(\frac{2m(2m+1)}{2}\right)^2 - 8\left(\frac{m(m+1)}{2}\right)$ , which further simplifies to $m^2(2m+1)^2-2m^2(m+1)^2$ .

Let's regroup. The equation we have right now is:

$\dfrac{m^2(2m+1)^2-2m^2(m+1)^2}{2m^2(m+1)^2} = \dfrac{721}{800}$

Next, we cross-multiply the fraction to get $800(m^2(2m+1)^2-2m^2(m+1)^2) = 1442m^2(m+1)^2$ . We simplify by expanding and combining like terms (not enough space here to show) to get a quartic polynomial, $158m^4-2884m^3-2242m^2=0$ . We solve this equation by dividing out $m^2$ (and note that $m = 0$ is also a solution to this equation, but it doesn't work for our problem). We get a quadratic for which the solutions are $\dfrac{-59}{79}$ , $0$ , and $19$ , but obviously only one of those can be our answer, and since $m$ must be a positive integer (in order for us to sum up cubes to it), we get the solution $\boxed{m = 19}$ . (phew, that was wordy.)

Look at this pattern: (1³/2³)=1/8, (1³+3³)/(2³+4³) = (1+6)/(8+10), (1³+3³+5³)/(2³+4³+6³) = (1+6+10)/(8+10+14) We must find the sum: [1+6+10+...+(4m+2)][/[8+10+14+18+...+(4m+6)]

By Sum of Arithmetic Progression:

= (2m²-1)/(2m²+4m+2) = 721/800

Solve a equation above and:

m =19

$\frac{1^3+3^3.....}{2^3+4^3....}=\frac{721}{800}$

Applying Componendo and Dividendo

$\frac{1^3+2^3.....}{2^3+4^3....}=\frac{1521}{800}$

$\frac{1^3+2^3.....}{2^3(1^3+2^3.....m^3)}=\frac{1521}{800}$

$\frac{(2m+1)^2}{(m+1)^2}=\frac{1521}{400}$

$\frac{(2m+1)}{(m+1)}=\frac{39}{20}$

$m=\boxed{19}$

The steps to do it are as follows:

Step 1 : Add $(2^3+4^3+6^3+...+(2m)^3)=x$ to the $N^r$ or make the $N^r$ as $(1^3+2^3+3^3+4^3+...+(2m-1)^3+(2m)^3)-(2^3+4^3+6^3+...+(2m)^3)$ .

We get the expression as: $\frac {(1^3+2^3+3^3+4^3+...+(2m-1)^3+(2m)^3)-x}{x}$

Step 2 : In $x$ we notice there is a factor common to every term which is $2^3$ . Taking common we get:

$x= 2^3(1^3+2^3+3^3+...+(m)^3)$

Step 3 : We use the formula $[\frac {n(n+1)}{2}]^{2}$ for finding the sum of cubes of first $n$ consecutive integers starting from $1^3$ ,then $2^3$ and so on.

Step 4 : We can easily evaluate $m$ now as we get a linear equation in variable $m$ ,solving which yields $m=\boxed{19}$ .

We must know if 1^3 + 2^3 + 3^3 + ... + n^3 = [n(n+1)/2]^2. The problem above can change into {1^3 + 2^3 + 3^3 + ... + (2m)^3} - {2^3 + 4^3 + 6^3 + ... + (2m)^3}/ 2^3 + 4^3 + 6^3 + ... + (2m)^3 = 721/800. ----> {[1^3 + 2^3 + 3^3 + ... + (2m)^3] / [2^3 + 4^3 + 6^3 + ... + (2m)^3]} - 1 = 721/800 ----> {[1^3 + 2^3 + 3^3 + ... + (2m)^3] / [2^3 + 4^3 + 6^3 + ... + (2m)^3]} = 1521/800 ----> {[1^3 + 2^3 + 3^3 + ... + (2m)^3] / 2^3[1^3 + 2^3 + 3^3 + ... + m^3]} = 1521/800 ----> {[1^3 + 2^3 + 3^3 + ... + (2m)^3] / 8[1^3 + 2^3 + 3^3 + ... + m^3]} = 1521/800 -----> {[1^3 + 2^3 + 3^3 + ... + (2m)^3] / [1^3 + 2^3 + 3^3 + ... + m^3]} = 1521/100 ----> {[2m/2.(2m+1)]^2}/{[m/2.(m+1)]^2} = 1521/100 -----> {2(2m+1)/(m+1)}^2 = (39/10)^2 ----> (4m+2)/(m+1) = 39/10 ----> 40m + 20 = 39m + 39 ----> m = 39 - 20 ---> m = 19. Answer : 19

Solution with the C Programming Language !

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#include <stdio.h>
#include <math.h>
#include <string.h>

int main()
{

double s=0,m=1,a=0,b=0,d=.90125;

/*As 721/800=.90125*/

while(m>-5)
{

a+=pow(m,3);
b+=pow(m+1,3);
s=a/b;
if(s==d){break;}
m+=2;
}

printf("The value of m = %.0lf",(m+1)/2);
printf("\n\n");

return 0;
}

@Hahn Lheem , nice problem!

One approach is to find the algebraic sums of the numerator and denominator separately and set it up in an equation and solve it. Here that turns out to be n^2(2n^2-1)/2n^2(1+n)^2 =721/800 Solving this we get two answers but discard the negative fraction one and keep the positive one which is 19.

The general term of the series in the numerator is $t_m = \left(2m - 1\right)^{3}$ . Hence, the numerator itself is given by:

$S_m = \sum\limits_{i = 1}^m \left(2m - 1\right)^{3}$

$\Rightarrow S_m = \sum\limits_{i = 1}^m \left(8m^{3} - 12m^{2} + 6m - 1\right)$

$\Rightarrow S_m = 8\sum\limits_{i = 1}^m m^3 - 12\sum\limits_{i = 1}^m m^2 + 6\sum\limits_{i = 1}^m m - \sum\limits_{i = 1}^m 1$

$\Rightarrow S_m = 8\left(\dfrac{m(m + 1)}{2}\right)^{2} - 12\left(\dfrac{m(m + 1)(2m + 1)}{6}\right) + 6\dfrac{m(m + 1)}{2} - m$

$\Longrightarrow S_m = m^{2}\left(2m^{2} - 1\right)$

Now, the general term of the series in the denominator is $t_m = \left(2m\right)^{3}$ . Hence, the denominator itself is given by:

$S_m = \sum\limits_{i = 1}^m \left(2m\right)^{3}$

$\Rightarrow S_m = 8\sum\limits_{i = 1}^m m^{3}$

$\Rightarrow S_m = 8\left(\dfrac{m(m+1)}{2}\right)^{2}$

$\Longrightarrow S_m = 2m^{2}\left(m+1\right)^{2}$

Dividing the numerator by the denominator, we get:

$\dfrac{m^{2}\left(2m^{2} - 1\right)}{2m^{2}\left(m+1\right)^{2}} = \dfrac{721}{800}$

$\Longrightarrow 79m^{2} - 1442m - 1121 = 0$

$\Rightarrow m = \boxed{19}$

...It's a bit long, innit? :) I'm certain to find a shorter solution the moment I publish this one.

Sum of cubes = (n(n+1)/2)^2;

sum of cubes of even numbers = 2*((n(n+1))^2);

sum of cubes of odd numbers = (n^2) ((2 (n^2))-1);

Note that we can get the numerator of the LHS and put it in summation notation to show that $\sum_{i=1}^m (2i-1)^3=\sum_{i=1}^m 8i^3 -12\sum_{i=1}^m i^2 +6\sum_{i=1}^m i -\sum_{i=1}^m 1$ . Using the summation formulas mentioned in the notes at the end we find that this is all equal to $2m^4-m^2$ . After applying the same method to the denominator and solving we find that $m=19$ .

Formulas:

$\sum_{i=1}^n k=kn$

$\sum_{i=1}^n i =\frac{i(i=1)}{2}$

$\sum_{i=1}^n i^2= \frac{i(i+1)(2i+1)}{6}$

$\sum_{i=1}^n i^3=\frac{i^2(i+1)^2}{4}$

(below formula is not used but is thrown in for good measure. Try proving all of these through induction if you can not already).

$\sum_{i=1}^n i^4= \frac{i(i+1)(2i+1)(3i^2+3i-1)}{30}$