MathLabs

Problem 2

Let a1,a2,a3,a4,a5a_1,a_2,a_3,a_4,a_5 be real numbers satisfying a1k2+1+a2k2+2+a3k2+3+a4k2+4+a5k2+5=1k2\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=1,2,3,4,5k=1,2,3,4,5. Find the value of a137+a238+a339+a440+a541\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 4 of 5: Divide by P
1−x2R(x)=−1120(x2−1)(x2−4)(x2−9)(x2−16)(x2−25)(x2+1)(x2+2)(x2+3)(x2+4)(x2+5)1-x^2R(x)=-\frac1{120}\frac{(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)}
Detailed analysis

Divide the identity by P(x). Since Q(x)=R(x)P(x), this gives the displayed formula for 1−x^2R(x).