The explicit formula \(x_n = 40n-n^2\) allows us to compute formulas for the previous term and the increment. Notice the use of parentheses to emphasize the role of grouped terms, especially when there will be a subtraction.

\begin{align*}
x_{n-1} &= 40(n-1) - (n-1)^2\\
& = 40(n-1) - (n-1)(n-1)\\
& = \big(40n - 40\big) - \big(n^2-2n+1\big)\\
& = 40n - 40 - n^2 + 2n - 1 \\
& = 42n - n^2 - 41
\end{align*}

\begin{align*}
\nabla x_n &= x_n - x_{n-1} \\
&= \big(40n - n^2\big) - \big(42n - n^2 - 41\big) \\
&= 40n - n^2 - 42n + n^2 + 41 \\
&= -2n + 41
\end{align*}

We can verify that our work looks correct by starting a table and checking whether the explicit formulas match what the terms should be.

\(n\) |
\(x_n\) |
\(x_{n-1}\) |
\(\nabla x_n\) |

0 |
\(40(0)-0^2 = 0\) |
undefined |
undefined |

1 |
\(40(1)-1^2 = 39\) |
\(42(1)-1^2 - 41 = 0\) |
\(-2(1)+41 = 39\) |

2 |
\(40(2)-2^2 = 76\) |
\(42(2)-2^2 - 41 = 39\) |
\(-2(2)+41 = 37\) |

Checking thes few values in the table gives us confidence that we did the algebra correctly. The formula for the previous sequence value is matching what we expect, as is the formula for the increment.

Now that we have a formula for the increments, we want to find the intervals where the increments are positive or negative. This corresponds to solving inequalities \(\nabla x_n \gt 0\) and \(\nabla x_n \lt 0\text{.}\) The increment is defined for index values \(n=1,2,\ldots\text{.}\)

The approach of solving an inequality by isolating the independent variable would go as follows. Start with the inequality in terms of the independent variable \(n\text{,}\) because we have an explicit definition for the sequence. To solve \(\nabla x_n \gt 0\text{,}\) we use balanced operations to create equivalent inequalities.

\begin{gather*}
\nabla x_n \gt 0\\
-2n+41 \gt 0 \\
-2n \gt -41 \\
\frac{-2n}{-2} \lt \frac{-41}{-2} \\
n \lt 20{\textstyle \frac{1}{2}}
\end{gather*}

When we divided both sides by \(-2\) (multiplied by \(-\frac{1}{2}\)), the equivalent relation showed a reversed inequality. The other inequality \(\nabla x_n \lt 0\) follows the same steps, resulting in the equivalent inequality

\begin{equation*}
\nabla x_n \lt 0 \quad \Leftrightarrow \quad n \gt 20{\textstyle \frac{1}{2}}\text{.}
\end{equation*}

The alternate approach involves solving the equation \(\nabla x_n = -2n+41 = 0\text{.}\) Solving the equation involves the same steps to give an equivalent equation

\begin{equation*}
\nabla x_n = 0 \quad \Leftrightarrow \quad n = 20\frac{1}{2} \text{.}
\end{equation*}

We now consider the intervals of values for \(n\) on either side of this value. The intervals are \(\{1,\ldots,20\}\) and \(\{21,\ldots,\infty\}\text{.}\) The principle for solving the inequality is to choose one value from each interval and use it to find the sign of \(\nabla x_n\text{.}\) For example, we can use \(n=10\) and \(n=25\text{.}\)

\begin{gather*}
\nabla x_{10} = -2(10)+41 = 21 \\
\nabla x_{25} = -2(25)+41 = -9
\end{gather*}

Both methods of solving the inequalities give the same intervals, which allow us to analyze the monotonicity of the sequence as shown in the table below.

Sign of \(\nabla x_n\) |
Monotonicity of \(x_n\) |

Positive on \(\{1,\ldots,20\}\) |
Increasing on \(\{0,\ldots,20\}\) |

Negative on \(\{21,\ldots,\infty\}\) |
Decreasing on \(\{20,\ldots,\infty\}\) |

Because \(x\) is increasing on \(\{0,\ldots,20\}\) and then decreasing on \(\{20,\ldots,\infty\}\text{,}\) \(x\) must have a maximum value at \(n=20\text{.}\) The value of the sequence at that index is

\begin{equation*}
x_{20} = 40(20) - 20^2 = 800-400 = 400.
\end{equation*}

To find concavity, we need to compute the second backward difference. This is computed like other backward differences.

\begin{align*}
\nabla^2 x_n &= \nabla x_n - \nabla x_{n-1}\\
&= \big(-2n+41\big) - \big(-2(n-1)+41\big)\\
&= \big(-2n+41\big) -\big(-2n+43\big)\\
&= -2n+41 + 2n - 43\\
&= -2
\end{align*}

The second backward difference is always negative, for \(n=2,3,\ldots\text{.}\) Consequently, \(x\) is concave down on \(\{0,\ldots,\infty\}\text{.}\)