Question

Find the indicated terms in each of the sequences whose $n^{th}$ terms are given by
(i) $a_n=\frac{n+2}{2n+3};a_7,a_9$
(ii) $a_n=(-1{)}^n{2}^{n+3}(n+1);a_5,a_8$
(iii) $a_n=2n^2-3n+1;a_5,a_7$
(iv) $a_n=(-1{)}^n(1-n+n^2);a_5,a_8$

Answer 1

(i) The $n$th term of the given sequence is $a_n=\frac{n+2}{2n+3}$.

To find the seventh and ninth term of the given sequence, substitute $n=7$ and $9$ in $a_n=\frac{n+2}{2n+3}$ as shown below:

$a_7=\frac{7+2}{(2\times 7)+3}=\frac{9}{14+3}=\frac{9}{17}$

$a_9=\frac{9+2}{(2\times 9)+3}=\frac{11}{18+3}=\frac{11}{21}$

Hence,$a_7=\frac{9}{17}$ and $a_9=\frac{11}{21}$.

(ii) The $n$th term of the given sequence is $a_n={(-1)}^n{2}^{n+3}(n+1)$.

To find the fifth and eighth term of the given sequence, substitute $n=5$ and $8$ in $a_n={(-1)}^n{2}^{n+3}(n+1)$ as shown below:

$a_5={(-1)}^5{2}^{5+3}(5+1)=-1{\times 2}^8\times 6=-1\times 256\times 6=-1536$

$a_8={(-1)}^8{2}^{8+3}(8+1)=-1{\times 2}^{11}\times 9=1\times 2048\times 9=18432$

Hence,$a_5=-1536$ and $a_8=18432$.

(iii) The $n$th term of the given sequence is $a_n={2n}^2-3n+1$.

To find the fifth and seventh term of the given sequence, substitute $n=5$ and $7$ in $a_n={2n}^2-3n+1$ as shown below:

$a_5={2\left(5\right)}^2-(3\times 5)+1=(2\times 25)-15+1=50-15+1=51-15=36$

$a_7={2\left(7\right)}^2-(3\times 7)+1=(2\times 49)-21+1=98-21+1=99-21=78$

Hence,$a_5=36$ and $a_7=78$.

(iv) The $n$th term of the given sequence is $a_n={(-1)}^n(1-n+n^2)$.

To find the fifth and eighth term of the given sequence, substitute $n=5$ and $8$ in $a_n={(-1)}^n(1-n+n^2)$ as shown below:

$a_5={(-1)}^5(1-5+5^2)=-1(1-5+25)=-1\times 21=-21$

$a_8={(-1)}^8(1-8+8^2)=1(1-8+64)=1\times 57=57$

Hence,$a_5=-21$ and $a_8=57$.