Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 160 / 1 / d / i

Table of contents
- Solution i

Solution

In a Young-diagram hook, the cell

(i, j)

has

λ_{i} - j

cells to its right and

λ_{j}^{'} - i

below it. Summing the arm lengths row by row and the leg lengths column by column gives

\sum_{(i, j) \in λ} h_{i, j} = \sum_{(i, j) \in λ} ((j - 1) + (i - 1) + 1) = \sum_{(i, j) \in λ} (i + j - 1) .

Adding the content

c_{i, j} = j - i

turns the summand on the right into

2 j - 1

. Since

\sum_{j = 1}^{m} (2 j - 1) = m^{2}

, summing each row proves

\sum_{(i, j) \in λ} (h_{i, j} + c_{i, j}) = \sum_{i} λ_{i}^{2} .

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