hello, solving the equation system $ \frac{a_1}{k^2+1}+\frac{a_2}{k^2+2}+\frac{a_3}{k^2+3}+\frac{a_4}{k^2+4}+\frac{a_5}{k^2+5}=\frac{1}{k^2}$ with $ k=1,2,3,4,5$ for $ a_1,a_2,a_3,a_4,a_5$ i have got $ a_1=\frac{11105}{72},a_2=-\frac{2673}{40},a_3=\frac{1862}{150},a_4=-\frac{1885}{18},a_5=\frac{1323}{40}$, from here we have $ \frac{a_1}{37}+\frac{a_2}{38}+\frac{a_3}{39}+\frac{a_4}{40}+\frac{a_5}{41}=\frac{187465}{6744582}$ Sonnhard.

how did u solve? just by substituting k values? or have u used any elegant method? otherwise it just becomes a problem of simple additon and multiplication

Reminds me of Cauchy Matrix... Can be restated in the following way: Compute $ (b_{6,2},\cdots,b_{6,6})M_{6,1}^{ - 1}(b_{1,1},\cdots,b_{5,1})^T$, where $ b_{i,j} = \frac {1}{i^2 + j - 1}$ and $ M_{i,j}$ represents a minor of a $ 6x6$ matrix with entries being $ b_{i,j}$.

Let \[ p(x) = x(x + 1)(x + 2) \dotsm (x + 5) \left( \frac{a_1}{x + 1} + \frac{a_2}{x + 2} + \dots + \frac{a_5}{x + 5} - \frac{1}{x} \right),\] so $ p(x)$ expands as a fifth degree polynomial. From the given equations, $ x = 1$, 4, 9, 16, and 25 are the roots of $ p(x)$, so \[ p(x) = c(x - 1)(x - 4)(x - 9)(x - 16)(x - 25)\] for some constant $ c$. We see that \begin{align*} p(x) &= (x + 1)(x + 2) \dotsm (x + 5) \left( \frac{a_1 x}{x + 1} + \frac{a_2 x}{x + 2} + \dots + \frac{a_5 x}{x + 5} - 1 \right) \\ &= c(x - 1)(x - 4)(x - 9)(x - 16)(x - 25). \end{align*} Taking $ x = 0$, we get $ -1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = (-1)(-4)(-9)(-16)(-25)c$, so $ c = 1/(1 \cdot 2 \cdot 3 \cdot 4 \cdot 5) = 1/120$. Hence, \begin{align*} &(x + 1)(x + 2) \dotsm (x + 5) \left( \frac{a_1 x}{x + 1} + \frac{a_2 x}{x + 2} + \dots + \frac{a_5 x}{x + 5} - 1 \right) \\ &= \frac{1}{120} (x - 1)(x - 4)(x - 9)(x - 16)(x - 25). \end{align*} Taking $ x = 36$, we get \[ 36 \cdot 37 \cdot 38 \cdot 39 \cdot 40 \cdot 41 \left( \frac{a_1}{37} + \frac{a_2}{38} + \dots + \frac{a_5}{41} - \frac{1}{36} \right) = \frac{35 \cdot 32 \cdot 27 \cdot 20 \cdot 11}{120},\] so \[ \frac{a_1}{37} + \frac{a_2}{38} + \dots + \frac{a_5}{41} = \frac{35 \cdot 32 \cdot 27 \cdot 20 \cdot 11}{120 \cdot 36 \cdot 37 \cdot 38 \cdot 39 \cdot 40 \cdot 41} + \frac{1}{36} = \frac{187465}{6744582}.\]

I think that I had almost exactly the same solution as nsato, but I made a small computation error during the test. How many points would I get for such a solution???

@nsato:your froof very nice

It's a very old idea. Let $f(x)=\sum_{i=1}^{5}{\frac{g(x)a_i}{x^2+i}}-\frac{g(x)}{x^2}$ where $g(x)=\prod_{i=0}^{5}{(x^2+i)}$. Note that $f(x)$ has $10$ roots,i.e.,$\pm{1},\pm{2},\pm{3},\pm{4},\pm{5}$ The degree of $f(x)$ is also $10$,so $f(x)=c(x^2-1)(x^2-4)(x^2-9)(x^2-16)(x^2-25)$ Setting $x=0$ one gets $-120=-g(0)=f(0)=-120^2 c \implies c=\frac{1}{120}$. Thus $f(x)$ is determined,and so now one can easily manipulate $\frac{f(6)}{37 \cdot 38 \cdot 39 \cdot 40 \cdot 41}$ which is our answer.

Help Why I'm wrong? f(k)= a1/(k^2+1) + a2/(k^2+2) + a3/(k^2+3) + a4/(k^2+4) + a5/(k^2+5)=1/k^2 g(k) fifth degree polynominal g(k)=c(k-1)(k-2)(k-3)(k-4)(k-5)=k^2.f(k)-1 g(0)=-c.5!=-1 c=1/5! g(6)=(1/5! ).5!=6^2.f(6)-1 =1 f(6)=2/36=1/18

we set $P(x)=x(x+1)(x+2)\cdots (x+5) \left (\frac{a_{1}}{x+1}+\frac{a_{2}}{x+2}+\cdots+\frac{a_{5}}{x+5}-\frac{1}{x}\right)$ clearly we can observe from problem statement that $1,4,9,16 ,25$ are roots hence we have $P(x)=k(x-1)(x-4)(x-9)(x-16)(x-25)$ now we can re write $P(x)$ as $P(x)=x(x+1)(x+2)\cdots (x+5) \left (\frac{a_{1}}{x+1}+\frac{a_{2}}{x+2}+\cdots+\frac{a_{5}}{x+5} \right)-(x+1)(x+2)\cdots (x+5)$ plugging $x=0$ we get $k=\frac{1}{120}$ hence we want to evaluate $P(x)$ at $36$ so doing so we get $ \frac{a_1}{37} + \frac{a_2}{38} + \dots + \frac{a_5}{41} = \frac{35 \cdot 32 \cdot 27 \cdot 20 \cdot 11}{120 \cdot 36 \cdot 37 \cdot 38 \cdot 39 \cdot 40 \cdot 41} + \frac{1}{36} = \boxed{\frac{187465}{6744582}}.$