Review question

# How are these recursively defined sequences related? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6555

## Solution

A list of real numbers $x_1$, $x_2$, $x_3$, … is defined by $x_1 = 1$, $x_2 = 3$ and then for $n\geq3$ by $x_n = 2x_{n−1} −x_{n−2} + 1.$

So, for example, $x_3 = 2x_2 −x_1 +1=2\times3−1+1=6$.

1. Find the values of $x_4$ and $x_5$.
Using the given formula, \begin{align*} x_4&=2x_3-x_2+1=2\times6-3+1=10,\\ x_5&=2x_4-x_3+1=2\times10-6+1=15. \end{align*}
1. Find values of real constants $A$, $B$, $C$ such that for $n = 1$, $2$, $3$, $\begin{equation*} x_n = A+Bn+Cn^2. \label{eq:starrepeat}\tag{*} \end{equation*}$

Putting $n = 1$, $2$, $3$, we have that $A$, $B$ and $C$ must satisfy $A+B+C=1,$ $A+2B+4C=3,$ and $A+3B+9C=6.$

Subtracting the second equation from the third gives us $B+5C=3$, and taking the first equation from the second gives $B+3C=2$.

These give us that $C=\tfrac{1}{2}$, $B=\tfrac{1}{2}$ and $A=0$.

So equation $\eqref{eq:starrepeat}$ becomes $x_n=\frac{1}{2}n(n+1).$

This is the formula for the sum of the first $n$ natural numbers.

1. Assuming that equation $\eqref{eq:starrepeat}$ holds true for all $n\geq1$, find the smallest $n$ such that $x_n\geq800$.

From above we know that $x_n\geq800$ requires $n(n+1)\geq1600$.

We know that $40^2=1600$, which means that $n=40$ will give us $n(n+1)=40\times41>1600$. To be sure it is the smallest $n$ we should check $n=39$. This leads to $n(n+1)=39\times40$ which is less than $1600$. So $n=40$ is the smallest value that gives $x_n\geq800$.

Alternatively we can solve $n(n+1) = 1600$, or $n^2+n-1600=0$.

Using the quadratic formula gives $n \approx 39.5$, and so $n(n+1)\geq1600$ for the first time when $n = 40$.

1. A second list of real numbers $y_1$, $y_2$, $y_3$, … is defined by $y_1 = 1$ and $y_n = y_{n−1} + 2n.$ Find, explaining your reasoning, a formula for $y_n$ which holds for $n\geq2$.
Let’s start by writing out the first few equations to deduce the $n$th term: \begin{align*} y_1&=1\\ y_2&=1+4\\ y_3&=1+4+6\\ \dots\\ y_n&=1+4+6+\cdots +2n. \end{align*}

Taking a closer look at the $n$th term, what we have is $y_n=1+2(2+3+4+\cdots+(n-1)+n) = 2(1+2+3+\cdots +n)-1.$

Looking at the right hand side of this, we see the sum of all the consecutive numbers from $1$ to $n$.

We know that this sum is equal to $\tfrac{1}{2}n(n+1)$, so our equation reads $y_n=n(n+1)-1.$

What is the approximate value of $\dfrac{x_n}{y_n}$ for large values of $n$?

We have $x_n=\frac{1}{2}n(n+1)$ and $y_n=n(n+1)-1.$ Notice that $y_n=2x_n-1$.

So the ratio becomes $\frac{x_n}{y_n} = \frac{2x_n}{2y_n} = \frac{y_n+1}{2y_n} = \frac{1}{2}+\frac{1}{2y_n}.$

For large values of $n$, $y_n$ is large, and so $\dfrac{1}{2y_n}$ is close to $0$.

This means that the ratio $\dfrac{x_n}{y_n}$ is approximately $\dfrac{1}{2}$.