Express the value in a single fraction

[anemone]

Let $x_1, x_2, x_3, x_4, x_5$ be real numbers satisfying the following equation:

$\dfrac{x_1}{m^2+1}+\dfrac{x_2}{m^2+2}+\dfrac{x_3}{m^2+3}+\dfrac{x_4}{m^2+4}+\dfrac{x_5}{m^2+5}= \dfrac{1}{m^2}$ for $m=1, 2, 3, 4, 5$.

Find the value of
$\dfrac{x_1}{37}+\dfrac{x_2}{38}+\dfrac{x_3}{39}+ \dfrac{x_4}{40}+\dfrac{x_5}{41}$

 

[eddybob123]

Spoiler

Fraction:$\frac{187465}{6744582}$

Continued fraction: $[0,35,1,44,111,2,2,12,4]$

Unit fraction expansion: $\frac{1}{36}+\frac{1}{58395}+\frac{1}{9724688047}+\frac{1}{283708672824669334580}$

 

[anemone]

Spoiler

Fraction:$\frac{187465}{6744582}$

Continued fraction: $[0,35,1,44,111,2,2,12,4]$

Unit fraction expansion: $\frac{1}{36}+\frac{1}{58395}+\frac{1}{9724688047}+\frac{1}{283708672824669334580}$

Thank you so much for participating, eddybob123.

Your answer is correct, but I hope you can share with me the method you used to solve this problem.

By sharing that means a brief explanation on the concept that you employed will suffice.

 

[anemone]

Let $x_1, x_2, x_3, x_4, x_5$ be real numbers satisfying the following equation:

$\dfrac{x_1}{m^2+1}+\dfrac{x_2}{m^2+2}+\dfrac{x_3}{m^2+3}+\dfrac{x_4}{m^2+4}+\dfrac{x_5}{m^2+5}= \dfrac{1}{m^2}$ for $m=1, 2, 3, 4, 5$.

Find the value of
$\dfrac{x_1}{37}+\dfrac{x_2}{38}+\dfrac{x_3}{39}+ \dfrac{x_4}{40}+\dfrac{x_5}{41}$

Solution provided by other:

Spoiler

Let $f(x)=\dfrac{x_1}{m^2+1}+\dfrac{x_2}{m^2+2}+\dfrac{x_3}{m^2+3}+\dfrac{x_4}{m^2+4}+\dfrac{x_5}{m^2+5}$, then

$f(\pm 1)=1$, $f(\pm 2)=\dfrac{1}{4}$, $f(\pm 3)=\dfrac{1}{9}$, $f(\pm 4)=\dfrac{1}{16}$, $f(\pm 5)=\dfrac{1}{25}$, and $f(6)$ is the value to be found.

Next, we let $g(x)=(x^2+1)(x^2+2)(x^2+3)(x^2+4)(x^2+5)$ and $h(x)=f(x)g(x)$.

Then for $m=\pm1, \pm2, \pm3, \pm4, \pm5$, we get $h(x)=f(x)g(x)=\dfrac{g(x)}{m^2}$, i.e. $g(x)-m^2h(x)=0$.

Since $g(x)-x^2h(x)$ is a polynomial of degree 10 with roots $\pm1, \pm2, \pm3, \pm4, \pm5$, we get

$g(x)-x^2h(x)=A(x^2-1)(x^2-4)(x^2-9)(x^2-16)(x^2-25)$ (*)

Putting $x=0$ we get $A=\dfrac{g(0)}{(-1)(-4)(-9)(-16)(-25)}=-\dfrac{1}{120}$.

Finally, dividing both sides of (*) by $g(x)$ gives

$\dfrac{g(x)-x^2h(x)}{g(x)}=-\dfrac{(x^2-1)(x^2-4)(x^2-9)(x^2-16)(x^2-25)}{120g(x)}$ (*)

$1-x^2\dfrac{h(x)}{g(x)}=1-x^2f(x)=-\dfrac{1}{120}\cdot\dfrac{(x^2-1)(x^2-4)(x^2-9)(x^2-16)(x^2-25)}{(x^2+1)(x^2+2)(x^2+3)(x^2+4)(x^2+5)}$ and hence

$1-36f(6)=\dfrac{35\cdot32\cdot27\cdot20\cdot11}{120 \cdot 37\cdot38\cdot39\cdot40\cdot41}=-\dfrac{231}{374699}$,

which implies $f(6)=\dfrac{187465}{6744582}$.