 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ib/paper-1/13b/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Only odd stationary states contribute, and for

j = 2 n + 1

their squared coefficients are

128/ [π^{2} (j^{2} - 16)^{2}]

. Since

E_{a v} / E_{1} = 16

, substitution gives

π^{2} = 8 \sum_{n \geq 0} \frac{( 2 n + 1 ) ^{2}}{[( 2 n + 1 ) ^{2} - 16 ] ^{2}},

B = 8

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