 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-1/8e/a/ii/solution

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

Since

λ I

commutes with

J_{d}

(λ I + J_{d})^{n} = k = 0 \sum m i n (n, d - 1) (k n) λ^{n - k} J_{d}^{k} .

Equivalently, its

k

th superdiagonal is constant with value

(k n) λ^{n - k}

for

0 \leq k < d

, with the convention that this value is zero when

k > n

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