Answer :

$a_n=n(n+2)$

Substituting $n=1,2,3,4$, and 5 , we obtain

$a_1=1(1+2)=3$

$a_2=2(2+2)=8$

$a_3=3(3+2)=15$

$a_4=4(4+2)=24$

$a_5=5(5+2)=35$

Therefore, the required terms are $3,8,15,24$, and 35 .

Answer :

$a_n=\dfrac{n}{n+1}$

Substituting $n=1,2,3,4,5$, we obtain

$a_1=\dfrac{1}{1+1}=\dfrac{1}{2}, a_2=\dfrac{2}{2+1}=\dfrac{2}{3}, a_3=\dfrac{3}{3+1}=\dfrac{3}{4}, a_4=\dfrac{4}{4+1}=\dfrac{4}{5}, a_5=\dfrac{5}{5+1}=\dfrac{5}{6}$

Therefore, the required terms are $\dfrac{1}{2}, \dfrac{2}{3}, \dfrac{3}{4}, \dfrac{4}{5}$, and $\dfrac{5}{6}$.

Answer :

$a_n=2^{n}$

Substituting $n=1,2,3,4,5$, we obtain

$a_1=2^{1}=2$

$a_2=2^{2}=4$

$a_3=2^{3}=8$

$a_4=2^{4}=16$

$a_5=2^{5}=32$

Therefore, the required terms are 2, 4, 8, 16, and 32 .

Answer :

Substituting $n=1,2,3,4,5$, we obtain

$a_1=\dfrac{2 \times 1-3}{6}=\dfrac{-1}{6}$

$a_2=\dfrac{2 \times 2-3}{6}=\dfrac{1}{6}$

$a_3=\dfrac{2 \times 3-3}{6}=\dfrac{3}{6}=\dfrac{1}{2}$

$a_4=\dfrac{2 \times 4-3}{6}=\dfrac{5}{6}$

$a_5=\dfrac{2 \times 5-3}{6}=\dfrac{7}{6}$

Therefore, the required terms are $\dfrac{-1}{6}, \dfrac{1}{6}, \dfrac{1}{2}, \dfrac{5}{6}$, and $\dfrac{7}{6}$

Answer :

Substituting $n=1,2,3,4,5$, we obtain

$a_1=(-1)^{1-1} 5^{1+1}=5^{2}=25$

$a_2=(-1)^{2-1} 5^{2+1}=-5^{3}=-125$

$a_3=(-1)^{3-1} 5^{3+1}=5^{4}=625$

$a_4=(-1)^{4-1} 5^{4+1}=-5^{5}=-3125$

$a^{5}=(-1)^{5-1} 5^{5+1}=5^{6}=15625$

Therefore, the required terms are $25, -125, 625, -3125,$ and $15625.$

Answer :

Substituting $n=1,2,3,4,5$, we obtain

$ \begin{aligned} & a_1=1 \cdot \dfrac{1^{2}+5}{4}=\dfrac{6}{4}=\dfrac{3}{2} \\ & a_2=2 \cdot \dfrac{2^{2}+5}{4}=2 \cdot \dfrac{9}{4}=\dfrac{9}{2} \\ & a_3=3 \cdot \dfrac{3^{2}+5}{4}=3 \cdot \dfrac{14}{4}=\dfrac{21}{2} \\ & a_4=4 \cdot \dfrac{4^{2}+5}{4}=21 \\ & a_5=5 \cdot \dfrac{5^{2}+5}{4}=5 \cdot \dfrac{30}{4}=\dfrac{75}{2} \end{aligned} $

Therefore, the required terms are $\dfrac{3}{2}, \dfrac{9}{2}, \dfrac{21}{2}, 21$, and $\dfrac{75}{2}$.

Answer :

Substituting $n=17$, we obtain

$a _{17}=4(17)-3=68-3=65$

Substituting $n=24$, we obtain

$a _{24}=4(24)-3=96-3=93$

Answer :

Substituting $n=7$, we obtain

$a_7=\dfrac{7^{2}}{2^7}=\dfrac{49}{128}$

Answer :

Substituting $n=9$, we obtain

$a_9=(-1)^{9-1}(9)^{3}=(9)^{3}=729$

Answer :

Substituting $n=20$, we obtain

$a _{20}=\dfrac{20(20-2)}{20+3}=\dfrac{20(18)}{23}=\dfrac{360}{23}$

Answer :

$a_1=3, a_n=3 a _{n-1}+2$ for all $n>1$

$\Rightarrow a_2=3 a_1+2=3(3)+2=11$

$a_3=3 a_2+2=3(11)+2=35$

$a_4=3 a_3+2=3(35)+2=107$

$a_5=3 a_4+2=3(107)+2=323$

Hence, the first five terms of the sequence are $3,11,35,107$, and 323 .

The corresponding series is $3+11+35+107+323+\ldots$

Answer :

$a_1=-1, a_n=\dfrac{a _{n-1}}{n}, n \geq 2$

$\Rightarrow a_2=\dfrac{a_1}{2}=\dfrac{-1}{2}$

$a_3=\dfrac{a_2}{3}=\dfrac{-1}{6}$

$a_4=\dfrac{a_3}{4}=\dfrac{-1}{24}$

$a_5=\dfrac{a_4}{5}=\dfrac{-1}{120}$

Hence, the first five terms of the sequence are

$ -1, \dfrac{-1}{2}, \dfrac{-1}{6}, \dfrac{-1}{24} \text{, and } \dfrac{-1}{120} $

The corresponding series is

$ (-1)+(\dfrac{-1}{2})+(\dfrac{-1}{6})+(\dfrac{-1}{24})+(\dfrac{-1}{120})+\ldots $

Answer :

$a_1=a_2=2, a_n=a _{n-1}-1, n>2$

$\Rightarrow a_3=a_2-1=2-1=1$

$a_4=a_3-1=1-1=0$

$a_5=a_4-1=0-1=-1$

Hence, the first five terms of the sequence are 2, 2, 1, 0 , and -1.

The corresponding series is $(2+2+1+0- 1)+\ldots$

Answer :

$1=a_1=a_2$

$a_n=a _{n-1}+a _{n-2}, n>2$

$\therefore a_3=a_2+a_1=1+1=2$

$a_4=a_3+a_2=2+1=3$

$a_5=a_4+a_3=3+2=5$

$a_6=a_5+a_4=5+3=8$

$\therefore$ For $n=1, \dfrac{a_n+1}{a_n}=\dfrac{a_2}{a_1}=\dfrac{1}{1}=1$

For $n=2, \dfrac{a_n+1}{a_n}=\dfrac{a_3}{a_2}=\dfrac{2}{1}=2$

For $n=3, \dfrac{a_n+1}{a_n}=\dfrac{a_4}{a_3}=\dfrac{3}{2}$

For $n=4, \dfrac{a_n+1}{a_n}=\dfrac{a_5}{a_4}=\dfrac{5}{3}$

For $n=5, \dfrac{a_n+1}{a_n}=\dfrac{a_6}{a_5}=\dfrac{8}{5}$