Using nth Terms of Sequences (AQA GCSE Further Maths): Revision Note

• If you are asked "which term", the question usually wants to know which value of n (e.g. n = 5, so the 5th term), not its value in the sequence

The n^th term of a sequence is given by $\frac{5-n}{2n^2+1}$

(a) Find the first three terms, simplifying your answers where possible.

The first three terms are when $n=1,n=2,n=3$ 

When $n=1$:   $\frac{5-1}{2{\left(1\right)}^2+1}=\frac{4}{2+1}=\frac{4}{3}$ 
 

When $n=2$:   $\frac{5-2}{2{\left(2\right)}^2+1}=\frac{3}{8+1}=\frac{3}{9}=\frac{1}{3}$
 

When $n=3$:   $\frac{5-3}{2{\left(3\right)}^2+1}=\frac{2}{18+1}=\frac{2}{19}$  

$\frac{\mathbf{4}}{\mathbf{3}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{,}\frac{\mathbf{2}}{\mathbf{19}}$

(b) Which term in the sequence is the first one to have a negative value? 
 
We can see from part (a) that the terms are decreasing, and getting closer to zero (and then negative numbers)

Let's find when the sequence is equal to zero, and then after this, the sequence will be negative 

$\frac{5-n}{2n^2+1}=0$ 

Multiplying both sides by the denominator and solving

$\begin{array}{rcl}5-n& =& 0\\ n& =& 5\end{array}$ 

So the 5^th term in the sequence is zero
As the sequence is decreasing, this means the 6^th term will be the first negative term
But we should substitute in $n=6$ to check this

$\frac{5-6}{2{\left(6\right)}^2+1}=\frac{-1}{72+1}=-\frac{1}{73}$ 

The 6^th term

Find the limiting values of the following sequences given by their n^th term formulae.

(a)$\begin{array}{c}\\ \end{array}$$\frac{5n+3}{2n-1}$
  

Divide the numerator and denominator by $n$ 

$\frac{5+{\displaystyle \frac{3}{n}}}{2-{\displaystyle \frac{1}{n}}}$ 

As $n$ tends towards infinity, $\frac{3}{n}$ tends towards 0, and $\frac{1}{n}$ tends towards 0
So the expression becomes 

$\frac{5+0}{2-0}$ 

So the limiting value is

$\frac{\mathbf{5}}{\mathbf{2}}$

(b)$\begin{array}{c}\\ \end{array}$$\frac{2-n^3}{n^3+4n^2-n}$ 
 
Divide the numerator and denominator by the highest power of $n$, which is $n^3$ 

$\frac{{\displaystyle \frac{2}{n^3}-1}}{1+{\displaystyle \frac{4}{n}}-{\displaystyle \frac{1}{n^2}}}$ 

As $n$ tends towards infinity, $\frac{2}{n^3}$ tends towards 0, $\frac{4}{n}$ tends towards 0, and $\frac{1}{n^2}$ tends towards 0
So the expression becomes

$\frac{0-1}{1+0-0}$ 

So the limiting value is

$\mathbf{-}\mathbf{1}$

(c)$\begin{array}{c}\\ \end{array}$$\frac{3n}{n^2+1}$  
 
Divide the numerator and denominator by the highest power of $n$, which is $n^2$ 

$\frac{{\displaystyle \frac{3}{n}}}{1+{\displaystyle \frac{1}{n^2}}}$ 

As $n$ tends towards infinity, $\frac{3}{n}$ tends towards 0, and $\frac{1}{n^2}$ tends towards 0
So the expression becomes

$\frac{0}{1+0}$ 

Notice that it is possible for the numerator to become zero
So the limiting value is

$\mathbf{0}$