A sequence of numbers \(H_1\), \(H_2\), \(H_3\), … is defined as follows: \(H_1=1\) and \(H_n=H_{n-1}+6(n-1)\) for \(n \ge 2\).

  1. Write down the numerical values of \(H_2\), \(H_3\) and \(H_4\).
\[\begin{align*} H_2&=H_1+6\times 1=1+6=7.\\ H_3&=H_2+6\times2=7+12=19.\\ H_4&=H_3+6\times 3=19+18=37. \end{align*}\]
  1. Find an expression for \(H_n\) in terms of \(n\).

We begin by working out what \(H_n\) is in terms of the earlier \(H_i\) values, and then use the formula for the sum of the integers from \(1\) to \(n-1\):

\[\begin{align*} H_n&=H_{n-1}+6(n-1)\\ &=H_{n-2}+6(n-2)+6(n-1)\\ &=\cdots\\ &=H_1+6\times1+6\times2+\cdots+6(n-2)+6(n-1)\\ &=H_1+6\sum_{i=1}^{n-1}i\\ &=1+6\times\frac{1}{2}n(n-1)\\ &=3n^2-3n+1. \end{align*}\]
  1. Prove that \[\sum_{i=1}^n H_i=n^3.\]
We have \[\begin{align*} \sum_{i=1}^n H_i&=\sum_{i=1}^n (3i^2-3i+1)\\ &=3\sum_{i=1}^n i^2 -3\sum_{i=1}^n i +\sum_{i=1}^n1\\ &=3\times \tfrac{1}{6}n(n+1)(2n+1)-3\times\tfrac{1}{2}n(n+1)+n\\ &=\tfrac{1}{2} n\bigl((n+1)(2n+1)-3(n+1)+2\bigr)\\ &=\tfrac{1}{2} n(2n^2+3n+1-3n-3+2)\\ &=\tfrac{1}{2} n(2n^2)\\ &=n^3 \end{align*}\]

as required.