Problem 2

Let

a_1,a_2,a_3,a_4,a_5

be real numbers satisfying

\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}=\frac1{k^2}

for

k=1,2,3,4,5

. Find the value of

\frac{a_1}{37}+\frac{a_2}{38}+\frac{a_3}{39}+\frac{a_4}{40}+\frac{a_5}{41}

(express the value in a single fraction).

Step 5 of 5: Evaluate at 6

Previous step Next step

1-36R(6)=-\frac{231}{374699}\Longrightarrow R(6)=\frac{187465}{6744582}

Detailed analysis

Substitute x=6. The numerator product is 35·32·27·20·11 and the denominator product is 37·38·39·40·41, so 1−36R(6)=−231/374699. Solving gives the required value R(6)=187465/6744582.