 Solution

ID: past-exam-of-the-geometry-course-of-the-university-of-oxford/2011/6/c/solution

x^{i} y^{j} z^{k}

with nonnegative integers

i + j + k = m

form a basis for the vector space of homogeneous polynomials of degree

m

. For each

i = 0, \dots, m

, there are

m - i + 1

(j, k)

. Hence

dim V_{m} = i = 0 \sum m (m - i + 1) = (2 m + 2) = \frac{( m + 1 ) ( m + 2 )}{2} .

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