The triangular numbers are Tk = 1 + 2 + 3 + … + k, so T₁ = 1, T₂ = 3, T₃ = 6, and so on…

(a) $T_{10} = \dfrac{10 \times 11}{2} = 55$.

(b) $\sum_{k=1}^{n} T_k = \dfrac12\left(\sum k^2 + \sum k\right) = \dfrac12\left(\dfrac{n(n + 1)(2n + 1)}{6} + \dfrac{n(n + 1)}{2}\right) = \dfrac{n(n + 1)(n + 2)}{6}$. For $n = 20$: $\dfrac{20 \times 21 \times 22}{6} = 1540$.

(c) $\left(\dfrac{15 \times 16}{2}\right)^2 = 120^2 = 14400$. The sum of the first $n$ cubes is the square of the $n$th triangular number.

(d) In pairs: $(1^2 - 2^2) + (3^2 - 4^2) + \dots = -(3 + 7 + 11 + \dots + 79)$, twenty terms of an AP. Their sum is $\dfrac{20}{2}(3 + 79) = 820$, so the answer is $-820$. (Neatly, it is $-(1 + 2 + \dots + 40) = -T_{40} = -820$, since $(2j - 1)^2 - (2j)^2 = -[(2j - 1) + 2j]$.)

(e) $\sum_{k=1}^{20} k^2 - \sum_{k=1}^{10} k^2 = \dfrac{20 \times 21 \times 41}{6} - \dfrac{10 \times 11 \times 21}{6} = 2870 - 385 = 2485$.