Review question

# Can we prove that $\sum_{i=1}^n H_i=n^3?$ Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9884

## Solution

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.